1,001 Calculus Practice Problems

Part II

The Answers

In this part …

Here you get answers and explanations for all 1,001 problems. As you read the solutions, you may realize that you need a little more help. Lucky for you, the For Dummies series (published by Wiley) offers several excellent resources. I recommend checking out the following titles, depending on your needs:

· Calculus For Dummies, Calculus Workbook For Dummies, and Calculus Essentials For Dummies, all by Mark Ryan

· Pre-Calculus For Dummies, by Yang Kuang and Elleyne Kase, and Pre-Calculus Workbook For Dummies, by Yang Kuang and Michelle Rose Gilman

· Trigonometry For Dummies and Trigonometry Workbook For Dummies, by Mary Jane Sterling

When you're ready to step up to more advanced calculus courses, you'll find the help you need in Calculus II For Dummies, by Mark Zegarelli.

Visit www.dummies.com for more information.

Chapter 16

Here are the answer explanations for all 1,001 practice problems.

1.

Get a common denominator of 12 and then perform the arithmetic in the numerator:

2.

Get a common denominator of 60 and then perform the arithmetic in the numerator:

3. 11

Write each factor as a fraction and then multiply across the numerator and denominator (cancel common factors as well!):

4.

Recall that to divide by a fraction, you multiply by its reciprocal: . After rewriting the fraction, cancel common factors before multiplying:

5.

To add the fractions, you need a common denominator. The least common multiple of the denominators is x²y, so multiply each fraction accordingly to get this denominator for each term. Then write the answer as a single fraction:

6.

You need a common denominator so you can subtract the fractions. The least common multiple of the denominators is (x – 1)(x + 1) = x² – 1, so multiply each fraction accordingly to get this denominator. Then perform the arithmetic in the numerator:

7.

Begin by factoring the expressions completely. Cancel any common factors and then simplify to get the answer:

8.

Recall that to divide by a fraction, you multiply by its reciprocal: . After rewriting the fraction, factor and cancel the common factors:

9.

Write each factor using positive exponents and then simplify. To start, notice that in the denominator becomes when it moves to the numerator:

10.

To simplify, cancel the common factors:

11.

Because the entire expression is being raised to the power of zero, the answer is 1:

There's no point in simplifying the expression inside the parentheses.

12.

Begin by factoring the numerator and then cancel the common factors:

13.

For this problem, you need to know the properties of exponents. The important properties to recall here are , , , , , and .

Write all factors using positive exponents and then use the properties of exponents to simplify the expression:

14.

Factor the number underneath the radical and then rewrite the radical using to simplify:

15.

Start by rewriting the expression so it has only one square root sign. Then reduce the fraction and use properties of radicals to simplify:

16.

Begin by writing the expression using a single square root sign. Then combine like bases and simplify:

17.

Begin by writing the expression using a single cube root sign. Combine the factors with like bases and then simplify:

18.

Simplify each root individually and then use properties of exponents to simplify further:

19.

Recall that . The expression becomes

20.

Recall that . Therefore, the expression becomes

21. y = 3x + 8

Recall that a function is one-to-one if it satisfies the requirement that if x₁ ≠ x², then f (x₁) ≠ f (x₂); that is, no two x coordinates have the same y value. If you consider the graph of such a function, no horizontal line would cross the graph in more than one place.

Of the given functions, only the linear function y = 3x + 8 passes the horizontal line test. This function is therefore one-to-one and has an inverse.

22. y = x² – 4, x ≥ 0

When determining whether a function has an inverse, consider the domain of the given function. Some functions don't pass the horizontal line test — and therefore don't have an inverse — unless the domain is restricted. The function y = x² – 4 doesn't pass the horizontal line test; however, if you restrict the domain to x ≥ 0, then y = x² – 4 does pass the horizontal line test (and therefore has an inverse), because you're getting only the right half of the parabola.

23. y = tan⁻¹ x

Of the given functions, only the function y = tan⁻¹ x passes the horizontal line test. Therefore, this function is one-to-one and has an inverse.

24.

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

Therefore, the inverse of f (x) = 4 – 5x is

25.

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse. You need to complete the square so you can isolate y. (To complete the square, consider the quadratic expression on the right side of the equal sign; take one-half of the coefficient of the term involving y, square it, and add that value to both sides of the equation.) Then factor the right side and use square roots to solve for y:

Notice that for the domain of the inverse to match the range of f (x), you have to keep the positive root. Therefore, the inverse of f (x) = x² – 4x, where x ≥ 2, is

26. ,

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

Therefore, the inverse of is

The domain of the inverse is x ≥ 0 because the range of f (x) is y ≥ 0 and the domain of is equal to the range of f (x).

27.

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

Therefore, the inverse of f (x) = 3x⁵ + 7 is

28.

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

Note that the range of is (–1, 1], so the domain of the inverse function is (–1, 1]. Therefore, the inverse of is

29.

First replace f (x) with y:

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

Therefore, the inverse of is

30. {(1, 0), (4, 3), (–6, 5)}

If the point (a, b) is on the graph of a one-to-one function, the point (b, a) is on the graph of the inverse function. Therefore, just switch the x and y coordinates to get that the set of points {(1, 0), (4, 3), (–6, 5)} belongs to the graph of f ⁻¹(x).

31. domain: (–1, 2); range: [–2, 4)

For a one-to-one function (which by definition has an inverse), the domain of f (x) becomes the range of f ⁻¹(x), and the range of f (x) becomes the domain of f ⁻¹(x). Therefore, f ⁻¹(x) has domain (–1, 2) and range [–2, 4).

32. g ⁻¹(x) = f ⁻¹(x) – c

Replacing x with (x + c) shifts the graph c units to the left, assuming that c > 0. If the point (a, b) belongs to the graph of f (x), then the point (a – c, b) belongs to the graph of f (x + c).

Now consider the inverse. The point (b, a) belongs to the graph of f ⁻¹(x), so the point (b, a – c) belongs to the graph of g ⁻¹(x). Therefore, g ⁻¹(x) is the graph of f ⁻¹(x) shifted down c units so that g ⁻¹(x) = f ⁻¹(x) – c.

The same argument applies if c < 0.

33. x = 2

Put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

34. x = –4

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

35.

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

36.

Put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x (that is, multiply both sides by the reciprocal of the coefficient) to get the solution:

37.

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

38. x = –3, 7

This quadratic factors without too much trouble using trial and error:

Setting each factor equal to zero gives you x – 7 = 0 so that x = 7 is a solution and x + 3 = 0 so that x = –3 is a solution.

39.

Complete the square to solve the quadratic equation. (To complete the square, consider the quadratic expression on the left side of the equal sign; take one-half of the coefficient of the term involving x, square it, and add that value to both sides of the equation.) Then factor the left side and use square roots to solve for x:

40.

Complete the square to solve the quadratic equation. Begin by moving the constant to the right side of the equal sign and then dividing both sides of the equation by 2:

Next, consider the quadratic expression on the left side of the equal sign; take one-half of the coefficient of the term involving x, square it, and add that value to both sides of the equation. Then factor the left side and use square roots to solve for x:

41.

Factoring by trial and error gives you

Setting each factor equal to zero gives you 2x – 1 = 0 so that is a solution and 3x + 4 = 0 so that is a solution.

42.

The equation 3x² + 4x – 2 = 0 doesn't factor nicely, so use the quadratic equation to solve for x. Here, a = 3, b = 4, and c = –2:

43.

The equation x¹⁰ + 7x⁵ + 10 = 0 isn't quadratic, but by using a substitution, you can produce a quadratic equation:

Now use the substitution y = x⁵ to form a quadratic equation that you can easily factor:

From y + 2 = 0, you have the solution y = –2_, and from y + 5 = 0, you have the solution y = –5_.

Now replace y using the original substitution and solve for x to get the final answer: x⁵ = –2 gives you _, and x⁵= –5 gives you .

44. x = 0, , 1

Begin by factoring out the greatest common factor, x². Then factor the remaining quadratic expression:

Next, set each factor equal to zero and solve for x: x² = 0 has the solution x = 0, 3x + 5 = 0 has the solution , and x – 1 = 0 has the solution x = 1.

45. no real solutions

Factoring this polynomial by trial and error gives you the following:

Setting the first factor equal to zero gives you x⁴ + 5 = 0 so that x⁴ = –5, which has no real solutions. Setting the second factor equal to zero gives you x⁴ + 7 = 0 so that x⁴ = –7, which also has no real solutions.

46. x = –1, 1

Factor the polynomial repeatedly to get the following:

Now set each factor equal to zero and solve for x. Setting the first factor equal to zero gives you x – 1 = 0 so that x = 1. The second factor gives you x + 1 = 0 so that x = –1. And the last factor gives you x² + 4 = 0 so that x² = –4, which has no real solutions.

47. x = –3, 3

Factor the polynomial repeatedly to get the following:

Now set each factor equal to zero and solve for x. Setting the first factor equal to zero gives you x – 3 = 0 so that x = 3. Setting the second factor equal to zero gives you x + 3 = 0 so that x = –3. The last factor gives you x² + 9 = 0, or x² = –9, which has no real solutions.

48. x = 1,

To solve an absolute value equation of the form , where b > 0, you must solve the two equations a = b and a = –b. So for the equation , you have to solve 5x – 7 = 2:

You also have to solve 5x – 7 = –2:

Therefore, the solutions are and x = 1.

49. no real solutions

To solve an absolute value equation of the form , where b > 0, you must solve the two equations a = b and a = –b. However, in this case, you have , which gives you . Because the absolute value of a number can't be negative, this equation has no solutions.

50. x = –3, 9

To solve an absolute value equation of the form , where b > 0, you must solve the two equations a = b and a = –b. So for the equation , you must solve x² – 6x = 27 and x² – 6x = –27.

To solve x² – 6x = 27, set the equation equal to zero and factor using trial and error:

Setting each factor equal to zero gives you x – 9 = 0, which has the solution x = 9, and x + 3 = 0, which has the solution x = –3.

Then set x² – 6x = –27 equal to zero, giving you x² – 6x + 27 = 0. This equation doesn't factor nicely, so use the quadratic equation, with a = 1, b = –6, and c = 27:

The number beneath the radical is negative, so this part of the absolute value has no real solutions.

Therefore, the only real solutions to are x = –3 and x = 9.

51. x = –3, 2

To solve an absolute value equation of the form , you must solve the two equations a = b and a = –b. So for the equation , you have to solve 15x – 5 = 35 – 5x and 15x – 5 = –(35 – 5x).

For the first equation, you have

And for the second equation, you have

Therefore, the solutions are x = –3 and x = 2.

52. x = –1

To solve a rational equation of the form , you need to solve the equation p(x) = 0. Therefore, to solve , you simply set the numerator equal to zero and solve for x:

53. x = ,

To solve the equation , first remove all fractions by multiplying both sides of the equation by the least common multiple of the denominators:

In this case, multiplying leaves you with a quadratic equation to solve:

Check the solutions to see whether they're extraneous (incorrect) answers. In this case, you can verify that both and satisfy the original equation by substituting these values into and checking that you get 1 on the left side of the equation.

54. –14

To solve an equation of the form , cross-multiply to produce the equation ad = bc:

After cross-multiplying, you're left with a quadratic equation, which then reduces to a linear equation:

You can verify that –14 is a solution of the original rational equation by substituting x = –14 into the equation and checking that you get the same value on both sides of the equation.

55. no real solutions

Begin by factoring all denominators. Then multiply each side of the equation by the least common multiple of the denominators to remove all fractions:

Then simplify:

Because x = 2 makes the original equation undefined, there are no real solutions.

56. (–4, 8)

To solve the inequality x² – 4x – 32 < 0, begin by solving the corresponding equation x² – 4x – 32 = 0. Then pick a point from each interval (determined by the solutions) to test in the original inequality.

Factoring x² – 4x – 32 = 0 gives you (x + 4)(x – 8) = 0. Setting the first factor equal to zero gives you x + 4 = 0, which has the solution x = –4, and setting the second factor equal to zero gives you x – 8 = 0, which has the solution x = 8.

Therefore, you need to pick a point from each of the intervals, (–∞, –4), (–4, 8), and (8, ∞), to test in the original inequality. Substitute each test number into the expression x² – 4x – 32 to see whether the answer is less than or greater than zero.

Using x = –10 to check the interval (–∞, –4) gives you

which is not less than zero and so doesn't satisfy the inequality.

Using x = 0 to check the interval (–4, 8) gives you

which is less than zero and does satisfy the inequality.

Using x = 10 to check the interval (8, ∞) gives you

which is not less than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the interval (–4, 8).

57.

To solve the inequality 2x⁴ + 2x³ ≥ 12x², begin by setting one side equal to zero. Solve the corresponding equation and then pick a point from each interval (determined by the solutions) to test in the inequality.

So from 2x⁴ + 2x³ ≥ 12x², you have 2x⁴ + 2x³ – 12x² ≥ 0. Factoring the corresponding equation so you can solve for x gives you

Set each factor equal to zero and solve for x, giving you the solutions x = 0,x = –3, and x = 2. These values are also solutions to the inequality.

Next, pick a test point from each of the intervals, (–∞, –3), (–3, 0), (0, 2), and (2, ∞), to see whether the answer is positive or negative. Using x = –10 to check the interval (–∞, –3) gives you

which is greater than zero.

Using x = –1 to check the interval (–3, 0) gives you

which is less than zero.

Using x = 1 to check the interval (0, 2) gives you

which is less than zero.

Finally, using x = 10 to check the interval (2, ∞) gives you

which is greater than zero.

Therefore, the solution set is .

Note that you could've divided the original inequality by 2 to simplify the initial inequality.

58.

To solve the rational inequality , first determine which values will make the numerator equal to zero and which values will make the denominator equal to zero. This helps you identify zeros (where the graph crosses the x-axis) and points where the function is not continuous.

From the numerator, you have x + 1 = 0, which has the solution x = –1. You also have x – 2 = 0, which has the solution x = 2. From the denominator, you have x + 3 = 0, which has the solution x = –3. Take a point from each interval determined by these solutions and test it in the original inequality.

You need to test a point from each of the intervals, (–∞, –3), (–3, –1), (–1, 2), and (2, ∞). Pick a point from each interval and substitute into the expression to see whether you get a value less than or greater than zero.

Using x = –10 to test the interval (–∞, –3) gives you

which is less than zero and so satisfies the inequality.

Using x = –2 to test the interval (–3, –1) gives you

which is greater than zero and doesn't satisfy the inequality.

Using x = 0 to test the interval (–1, 2) gives you

which is less than zero and so satisfies the inequality.

Using x = 3 to test the interval (2, ∞) gives you

which is greater than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the interval (–∞, –3) and (–1, 2).

59. (1, 3)

Begin by putting the rational inequality into the form by putting all terms on one side of the inequality and then getting common denominators:

Next, set the numerator equal to zero and solve for x. Setting the numerator equal to zero gives you a quadratic equation that you can factor by trial and error:

The solutions are and x = 3.

Also set each factor from the denominator equal to zero and solve for x. This gives you both (x – 1) = 0, which has the solution x = 1, and (x + 1) = 0, which has the solution x = –1.

Next, take a point from each of the intervals, (–∞, –1), , , (1, 3), and (3, ∞), and test it in the expression to see whether the answeris positive or negative; or equivalently, you can use the expression .

Using x = –10 to test the interval (–∞, 1) gives you

which is less than zero and so doesn't satisfy the inequality.

Using to test the interval gives you

which is greater than zero and so satisfies the inequality.

Using x = 0 to test the interval gives you

which is less than zero and so doesn't satisfy the inequality.

Using x = 2 to test the interval (1, 3) gives you

which is greater than zero and so satisfies the inequality.

And using x = 10 to test the interval (3, ∞) gives you

which is less than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the intervals and (1, 3).

60.

To solve an absolute inequality of the form , where b > 0, solve the compound inequality –b < a < b. So for the inequality , you have

The solution set is .

61. (–∞, 1)

To solve an absolute value inequality of the form , where b > 0, you have to solve the corresponding inequalities a > b and a < –b. Therefore, for the inequality , you have to solve 5x – 7 > 2 and 5x – 7 < –2. For the first inequality, you have

And for the second inequality, you have

Therefore, the solution set is the intervals (–∞, 1) and .

62.

To solve an absolute inequality of the form , where b > 0, solve the compound inequality –b ≤ a ≤ b. For the inequality , you have

Therefore, the solution is the interval .

63.

The slope of a line that goes through the points (x₁, y₁) and (x₂, y₂) is given by . Therefore, the slope of the line that passes through (1, 2) and (5, 9) is

64. y = 4x + 5

The graph of y = mx + b is a line with slope of m and a y-intercept at (0, b). Therefore, using m = 4 and b = 5 gives you the equation y = 4x + 5.

65.

The equation of a line that goes through the point (x₁, y₁) and has a slope of m is given by the point-slope formula: y – y₁ = m(x – x₁).

The slope of the line that goes through (–2, 3) and (4, 8) is

Now use the slope and the point (4, 8) in the point-slope formula:

66.

The equation of a line that goes through the point (x₁, y₁) and has a slope of m is given by the point-slope formula: y – y₁ = m(x – x₁).

The two lines are parallel, so the slopes are the same. Therefore, the slope of the line you're trying to find is .

Using the slope and the point (1, 5) in the point-slope formula gives you

67.

The equation of a line that goes through the point (x₁, y₁) and has a slope of m is given by the point-slope formula: y – y₁ = m(x – x₁).

The slope of the line that goes through the points (–6, 2) and (3, –4) is

The slopes of perpendicular lines are opposite reciprocals of each other, so the slope of the line you want to find is (you flip the fraction and change the sign).

Using the slope and the point (3, –4) in the point-slope formula gives you

Here's what the perpendicular lines look like:

68.

Stretching the graph of vertically by a factor of 2 produces the equation . To shift the graph of to the right 3 units, replace x with (x – 3) to get . Last, to move the graph of up 4 units, add 4 to the right side to get .

69.

The vertex form of a parabola is given by y = a(x – h)² + k, with the vertex at the point (h, k). Using the vertex (–2, –4) gives you the following equation:

Now use the point (0, 2) to solve for a:

Therefore, the equation of the parabola is

70. y = –(x + 1)² + 6

The vertex form of a parabola is given by y = a(x – h)² + k, with the vertex at the point (h, k). Using the vertex (–1, 6), you have the following equation:

Now use the point (1, 2) to solve for a:

Therefore, the vertex form of the parabola is

Here's the graph of y = –(x + 1)² + 6:

71. y = 3(x – 5)² + 2

To translate the parabola 6 units to the right, you replace x with the quantity (x – 6). And to translate the parabola 2 units down, you subtract 2 from the original expression for y:

Another option is simply to count and determine that the new vertex is at (5, 2); then use the vertex form y = a(x – h)² + k, where the vertex is at the point (h, k).

72. y = e^−2x – 5

To compress the graph of y = e^x horizontally by a factor of 2, replace x with 2x to get y = e^2x. To reflect the graph of y = e^2x across the y-axis, replace x with –x to get y = e^2(–x), or y = e^−2x. Last, to shift the graph of y = e^−2x down 5 units, subtract 5 from the right side of the equation to get y = e^−2x – 5.

Here's the graph of y = e^−2x – 5:

73.

To stretch the graph of horizontally by a factor of 5, replace x with to get . Next, to reflect the graph of across the x-axis, multiply the right side of the equation by –1 to get . Last, to shift the graph of up 3 units, add 3 to the right side of the equation to get

74.

The polynomial has x intercepts at x = –4, x = –2, and x = 1, so you know that the polynomial has the factors (x + 4), (x + 2), and (x – 1). Therefore, you can write

Use the point (0, 3) to solve for a:

Now you can enter the value of a in the equation of the polynomial and simplify:

Here's the graph of :

75.

The polynomial is fourth-degree and has the x-intercepts at x = –1, 2, and 3, where 3 is a repeated root. Therefore, the polynomial has the factors (x + 1), (x – 2), and (x – 3)², so you can write

Use the point (1, 4) to solve for a:

Now you can enter the value of a in the equation of the polynomial and simplify:

76.

The parabola has x-intercepts at x = –4 and x = 6, so you know that (x + 4) and (x – 6) are factors of the parabola. Therefore, you can write

Use the point (0, 8) to solve for a:

Now you can enter the value of a in the equation of the parabola and simplify:

77.

The parabola has x-intercepts at x = –8 and x = –2, so you know that (x + 8) and (x + 2) are factors of the parabola. Therefore, you can write

Use the point (–4, –12) to solve for a:

Now you can enter the value of a in the equation of the parabola and simplify:

78. domain: [–2, ∞); range: [1, ∞)

The function is continuous on its domain.

The lowest x coordinate occurs at the point (–2, 3), and then the graph extends indefinitely to the right; therefore, the domain is [–2, ∞). The lowest y coordinate occurs at the point (–1, 1), and then the graph extends indefinitely upward; therefore, the range is [1, ∞).

Note that the brackets in the interval notation indicate that the value –2 is included in the domain and that 1 is included in the range.

79. domain: [–5, 3]; range: [–2, 2]

Notice that the function is continuous on its domain.

The smallest x coordinate occurs at the point (–5, 0), and the largest x coordinate occurs at the point (3, 0); therefore, the domain is [–5, 3]. The smallest y coordinate occurs at the point (1, –2), and the largest y coordinate occurs at the point (–3, 2); therefore, the range is [–2, 2].

Note that the brackets in the interval notation indicate that the values –5 and 3 are included in the domain and that the values –2 and 2 are included in the range.

80. domain: ; range:

Notice that the function is continuous on its domain.

For the portion of the graph that's to the left of the y-axis, there's no largest x value due to the hole at (–1, –1); instead, the x values get arbitrarily close to –1. The graph then extends indefinitely to the left so that the domain for the left side of the graph is (–∞, –1). There's also no largest y value due to the hole at (–1, –1); the y values get arbitrarily close to –1. The graph then decreases as it approaches the horizontal asymptote at y = –3, so the range for this portion of the graph is (–3, –1).

For the portion of the graph that's to the right of the y-axis, the smallest x value is at the point (2, 2). The graph then extends indefinitely to the right, so the domain for the right side of the graph is [2, ∞). The smallest y value also occurs at the point (2, 2), and then the graph increases as it approaches the horizontal asymptote at y = 4; therefore, the range of this portion of the graph is [2, 4).

Putting the two parts together, the domain of the function is , and the range is .

81. ,

To determine the end behavior of a polynomial, you just have to determine the end behavior of the highest-powered term.

As x approaches –∞, you have

Note that even though the limit is approaching negative infinity, the limit is still positive due to the even exponent.

As x approaches ∞, you have

82. ,

To determine the end behavior of a polynomial, determine the end behavior of the highest-powered term.

As x approaches –∞, you have

Note that the limit is positive because a negative number raised to an odd power is negative, but after multiplying by –7, the answer becomes positive.

As x approaches ∞, you have

Likewise, the limit here is negative because a positive number raised to an odd power (or any power for that matter!) is positive, but after multiplying by –7, the answer becomes negative.

83. 3x + 12

Remove the parentheses and combine like terms:

84. 3x – 2

Remove the parentheses and combine like terms:

85. x³ – x² + 2x + 14

Remove the parentheses and combine like terms:

86. –2x⁴ + 3x + 8

Remove the parentheses and combine like terms:

87. x⁴ + 5x³ – 3x²

Remove the parentheses and combine like terms:

88. 3x – 7

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

89. 6x² – 5x + 5

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

90. 4x³ + 5x² – 8x + 14

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

91. 2x² + 3x + 1

Begin by removing the parentheses, being careful to distribute the –1 to the second and third terms. Then collect like terms to simplify:

92. 10x⁴ – 7x³ – 9x² – 8x + 10

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

93. 5x³ – 15x²

Distribute to get

94. 3x² + 7x – 20

Distribute each term in the first factor to each term in the second factor. Then collect like terms:

95. x²y – xy² + 6xy

Multiply each term in the first factor by xy:

96. 2x³ – 3x² + 9x – 4

Distribute each term in the first factor to each term in the second factor and then collect like terms:

97.

Begin by distributing –x to each term in the second factor:

Next, multiply each term in the first factor by each term in the second factor:

98.

Using polynomial long division gives you

Remember to put the remainder over the divisor when writing the answer:

99.

Using polynomial long division gives you

When writing the answer, put the remainder over the divisor:

100.

First add a placeholder for the missing x² term in the numerator. Rewriting x³ – 2x + 6 as x³ + 0x² – 2x + 6 will make all the like terms line up when you do the long division, making the subtraction a bit easier to follow.

Then use polynomial long division:

When you write the answer, put the remainder over the divisor:

101.

Begin by filling in all the missing terms so that everything will line up when you perform the long division. Here, add 0x³ and 0x as placeholders in the numerator and put 0x in the denominator:

Then use polynomial long division:

Put the remainder over the divisor when writing your answer:

102.

Begin by filling in all the missing terms so that everything will line up when you perform the long division:

Using polynomial long division gives you

Then write your answer, putting the remainder over the divisor:

103. ; ;

When considering the sides of the right triangle, the values of the trigonometric functions are given by , , and ; therefore, , , and .

104. ; ;

When considering the sides of the right triangle, the values of the trigonometric functions are given by , , and ; therefore, , , and .

105.

You know the value of , so if you can find the value of , you can evaluate using . To find , use the identity :

Next, take the square root of both sides, keeping the negative solution for cosine because :

Therefore, using , you have

106.

To find the value of , you can find the value of and then use . Use the identity to solve for :

Next, take the square root of both sides, keeping the negative solution for sine because :

Therefore, using , you have

107.

To find the value of , you can use the identity . Notice that because and , angle must be in the second quadrant. Using , you can make a right triangle and find the missing side using the Pythagorean theorem. When making the triangle, you can neglect the negative sign:

Because is in the second quadrant, you have and . Enter these values in the identity and solve:

108.

To find the value of , you can use the identity . Using you can make a right triangle and find the missing side using the Pythagorean theorem. When making the triangle, you can neglect the negative sign:

Now use the identity . The sine is negative, so you have :

109. rad

Because 180° = π rad, you have , so multiply the number of degrees by this value:

110. rad

Because 180° = π rad, you have , so multiply the number of degrees by this value:

111. rad

Because 180° = π rad, you have , so multiply the number of degrees by this value:

112. rad

Because 180° = π rad, you have , so multiply the number of degrees by this value:

113. 210°

Because π rad = 180°, you have , so multiply the number of radians by this value:

114. 165°

Because π rad = 180°, you have , so multiply the number of radians by this value:

115. –108°

Because π rad = 180°, you have , so multiply the number of radians by this value:

116. –630°

Because π rad = 180°, you have , so multiply the number of radians by this value:

117.

Because is in Quadrant II and the angle is measured counterclockwise, you have . Therefore, is the most appropriate choice.

118.

Because is in Quadrant III and the angle is measured clockwise, you have . Therefore, is the most appropriate choice.

119.

Because is in Quadrant IV and the angle is measured counterclockwise, you have . Therefore, is the most appropriate choice.

120. ; ;

The given angle measure is . Using the first quadrant of the unit circle, you have the following:

121. ; ;

The given angle measure is . Using the second quadrant of the unit circle, you have the following:

122. ; ;

Use the third quadrant of the unit circle. Note that because the angle touches the unit circle in the same place as the angle , you have the following:

123. ; ;

Use the third quadrant of the unit circle. The angle touches the unit circle in the same place as the angle , so you have the following:

124. ; ;

Because the angle intersects the unit circle at the point (–1, 0), you have the following:

125.

Simply rewrite cotangent and simplify:

126. sin x tan x

Begin by writing sec x as . Then get common denominators and simplify:

127. 1 + sin 2x

Expand the given expression and use the identity sin² x + cos² x = 1 along with 2 sin x cos x = sin(2x):

128. sin x

Use the identity sin(A – B) = sin A cos B – cos A sin B on the expression sin(π – x):

129. cos x

Use the identity cos A cos B + sin A sin B = cos(A – B) to get

Note that you can also use the following:

130.

First get common denominators:

Then use the identity sin² x = 1 – cos² x in the denominator:

131. csc x + cot x

Begin by multiplying the numerator and denominator by the conjugate of the denominator, 1 + cos x. Then start to simplify:

Continue, using the identity 1 – cos² x = sin² x to simplify the denominator:

132.

Use the identity cos(A + B) = cos A cos B – sin A sin B:

Then use the identities and and simplify:

133. ,

Solve for sin x:

The solutions are x = and in the interval [0, 2π].

134. 0, π, 2π

Make one side of the equation zero, use the identity , and then factor:

Setting the first factor equal to zero gives you sin x = 0, which has the solutions x = 0, π, and 2π. Setting the second factor equal to zero gives you so that cos x = 1, which has the solutions x = 0 and 2π.

135. , π,

Factor the equation:

Setting the first factor equal to zero gives you 2 cos x – 1 = 0 so that , which has the solutions and . Setting the second factor equal to zero gives you cos x + 1 = 0 so that cos x = –1, which has the solution x = π.

136. , , ,

To solve the equation , you must find the solutions to two equations: tan x = 1 and tan x = –1. For the equation tan x = 1, you have the solutions and . For the equation tan x = –1, you have the solutions and .

137. ,

Factor the equation 2 sin² x – 5 sin x – 3 = 0 to get

Setting the first factor equal to zero gives you 2 sin x + 1 = 0 so that , which has the solutions and . Setting the second factor equal to zero gives you sin x – 3 = 0 so that sin x = 3. Because 3 is outside the range of the sine function, this equation from the second factor has no solution.

138. ,

Begin by making one side of the equation equal to zero. Then use the identity and factor:

Setting the first factor equal to zero gives you cos x = 0, which has the solutions x = and . Setting the second factor equal to zero gives you so that sin x = 1, which has the solution .

139. , , ,

Begin by making the substitution 2x = y. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π and that 0 ≤ y ≤ 4π.

Solving the equation gives you the solutions , , , and in the interval [0, 4π].

Next, take each solution, set it equal to 2x, and solve for x: so that ; so that ; so that ; and so that .

140. , , ,

Begin by making one side of the equation zero. Then use the identity sin 2x = 2 sin x cos x and factor:

Setting the first factor equal to zero gives you cos x = 0, which has the solutions and . Setting the second factor equal to zero gives you 2 sin x – 1 = 0 so that , which has the solutions and .

141. ,

Begin by using the identity and then factor:

Set each factor equal to zero and solve for x. The equation cos x = 0 has the solutions and , and the equation 1 + sin x = 0 gives you sin x = –1, which has the solution .

142. , ,

Use the identity cos 2x = 2 cos² x – 1, make one side of the equation zero, and factor:

Set each factor equal to zero and solve for x. Setting the first factor equal to zerogives you 2 cos x + 1 = 0 so that , which has the solutions and . Setting the second factor equal to zero gives you cos x + 1 = 0 so that cos x = –1, which has the solution x = π.

143. , , , , ,

Begin by making the substitution y = 3x. Because 0 ≤ x ≤ 2π, you have 0 ≤ 3x ≤ 6π so that 0 ≤ y ≤ 6π. Finding all solutions of the equation tan y = –1 in the interval [0, 6π] gives you , , , , , and .

Next, take each solution, set it equal to 3x, and solve for x: so that ; so that ; so that ; so that ; so that ; and so that .

144. , , ,

Begin by making one side of the equation equal to zero and then factor:

Next, make the substitution y = 2x:

Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π so that 0 ≤ y ≤ 4π.

Setting the first factor equal to zero gives you cos y = 0, which has the solutions , , , and . Take each solution, set it equal to 2x, and solve for x: so that ; so that ; so that ; and so that .

Proceed in a similar manner for the second factor: so that sin y = 1, which has the solutions and . Now take each solution, set it equal to 2x, and solve for x: so that , and so that .

145. amplitude: ; period: 2π; phase shift: ; midline y = 0

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. Writing gives you the amplitude as , the period as , the phase shift as , and the midline as y = 0.

146. amplitude: ; period: 2; phase shift: ; midline y = 0

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. Writing gives you the amplitude as , the period as , the phase shift as , and the midline as y = 0.

147. amplitude: 3; period: 2; phase shift: ; midline y = 2

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. Writing gives you the amplitude as , the period as , the phase shift as , and the midline as y = 2.

148. amplitude: 1; period: 4π; phase shift: –π; midline

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. Writing gives you the amplitude as , the period as , the phase shift as –π, and the midline as .

149. f (x) = 2 sin(2x)

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a sine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 0 is the midline of the function so that D = 0. Last, the amplitude is 2, so you can use A = 2 because the function increases for values of x that are slightly larger than zero. Therefore, the function describes the given graph. Note that you can also use other functions to describe this graph.

150. f (x) = 2 cos(2x)

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a cosine function to describe the graph, there's no phase shift; therefore, you can use C = 0. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2; you can use A = 2 because the function decreases for values of x that are slightly larger than zero. Therefore, the function f (x) = 2 cos(2x) describes the given graph. Note that you can also use other functions to describe this graph.

151.

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of 4π, so one possible value of B is . If you use a cosine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than zero. Therefore, the function describes the given graph. Note that you can also use other functions to describe this graph.

152. f (x) = 2 cos(2x) + 1

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a cosine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 1 is the midline of the function, so D = 1. Last, the amplitude is 2, so you can use A = 2 because the function decreases for values of x that are slightly larger than zero. Therefore, the function f (x) = 2 cos(2x) + 1 describes the given graph. Note that you can also use other functions to describe this graph.

153. For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of 4π, so one possible value of B is . Think of the graph as a cosine graph that's flipped about the x-axis and shifted to the right; that means there's a phase shift of to the right, so . The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than . Therefore, the following function describes the graph:

Note that you can also use other functions to describe this graph.

154.

For the function , the amplitude is , the period is , the phase shift is , and the midline is y = D. The function has a period of 4π, so one possible value of B is . Think of the graph as a cosine graph that's flipped about the x-axis and shifted to the right and down; there's a phase shift of to the right so that . The line y = –1 is the midline of the function, so D = –1. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than . Therefore, the following function describes the given graph.

Note that you can also use other functions to describe this graph.

155.

To evaluate , find the solution of , where . Because , you have .

156.

To evaluate arctan(–1) = x, find the solution of –1 = tan x, where . Because , you have .

157.

To evaluate , first find the value of . To evaluate , where , find the solution of . Because , you have . Therefore, .

158.

To evaluate , first find the value of . To evaluate , where , find the solution of . Because , you have . Therefore, .

159.

The value of probably isn't something you've memorized, so to evaluate , you can create a right triangle.

Let so that . Using , create the right triangle; then use the Pythagorean theorem to find the missing side:

By the substitution, you have , and from the right triangle, you have .

160.

To evaluate , first create two right triangles using the substitutions tan⁻¹(2) = α and tan⁻¹(3) = β.

To make the first right triangle, use tan⁻¹(2) = α. You know that 2 = tan α, and you can find the missing side of the first right triangle using the Pythagorean theorem:

As for the second right triangle, because tan⁻¹(3) = β, you know that 3 = tan β. Again, you can use the Pythagorean theorem to find the missing side of the right triangle:

The substitutions give you . Using a trigonometric identity, you know that sin(α + β) = sin α cos β + cos α sin β. From the right triangles, you can read off each of the values to get the following:

161. x = 0.412, π – 0.412

To solve sin x = 0.4 for x over the interval [0, 2π], begin by taking the inverse sine of both sides:

The other solution belongs to Quadrant II because the sine function also has positive values there:

162. x = 2.465, 2π – 2.456

To solve cos x = –0.78 for x over the interval [0, 2π], begin by taking the inverse cosine of both sides:

The other solution belongs to Quadrant III because the cosine function also has negative values there:

163. x = 0.322, , π + 0.322,

To solve 5 sin(2x) + 1 = 4 for x over the interval [0, 2π], begin by isolating the term involving sine:

You can also use the substitution y = 2x to help simplify. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π so that 0 ≤ y ≤ 4π. Use the substitution and take the inverse sine of both sides:

It follows that in the interval [0, 4π], 2π + 0.644 is also a solution because when you add 2π, the resulting angle lies at the same place on the unit circle.

Likewise, there's a solution to the equation in the second quadrant because the function also has positive values there, namely π – 0.644; adding 2π gives you the other solution, 3π – 0.644. Therefore, y = 0.644, π – 0.644, 2π + 0.644, and 3π – 0.644 are all solutions.

Last, substitute 2x into the equations and divide to get the solutions for x:

The solutions are x = 0.322, , π + 0.322, and .

164. x = 0.321, , , , , 2π – 0.321

To solve 7 cos(3x) – 1 = 3 for x over the interval [0, 2π], first isolate the term involving cosine:

You can also use the substitution y = 3x to simplify the equation. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 3x ≤ 6π so that 0 ≤ y ≤ 6π. Use the substitution and take the inverse cosine of both sides:

It follows that in the interval [0, 6π], y = 2π + 0.963 and y = 4π + 0.963 are also solutions because adding multiples of 2π makes the resulting angles fall at the same places on the unit circle.

Likewise, there's a solution to the equation in the fourth quadrant because cosine also has positive values there, namely y = 2π – 0.963. Because y = 2π – 0.963 is a solution, it follows that y = 4π – 0.963 and y = 6π – 0.963 are also solutions.

Therefore, you have y = 0.963, 2π – 0.963, 2π + 0.963, 4π – 0.963, 4π + 0.963, and 6π – 0.963 as solutions to . Last, substitute 3x into the equations and divide to solve for x:

Therefore, the solutions are x = 0.321, , , , , and 2π – 0.321.

165. x = π + 0.887, 2π – 0.887

To solve 2 sin² x + 8 sin x + 5 = 0 for x over the interval [0, 2π], first use the quadratic formula:

Simplifying gives you and .

You now need to find solutions to sin x = –0.775 and sin x = –3.225. Notice that sin x = –3.225 has no solutions because –3.225 is outside the range of the sine function.

To solve sin x = –0.775, take the inverse sine of both sides:

Note that this solution isn't in the desired interval, [0, 2π]. The solutions in the given interval belong to Quadrants III and IV, respectively, because in those quadrants, the sine function has negative values; those solutions are x = π + 0.887 and x = 2π – 0.887.

166. x = π – 0.322, 2π – 0.322, ,

To solve 3 sec² x + 4 tan x = 2 for x over the interval [0, 2π], begin by using the identity sec² x = 1 + tan² x and make one side of the equation equal to zero:

Next, use the quadratic formula to find tan x:

Therefore, tan x = –1 and .

The solutions to the equation tan x = –1 are and . To solve , take the inverse tangent of both sides:

Note that this solution isn't in the given interval. The solutions that are in the given interval and belong to Quadrants II and IV (where the tangent function is negative) are x = π – 0.322 and x = 2π – 0.322.

167. 1

As x approaches 3 from the left, the y values approach 1 so that .

168. 3

As x approaches 3 from the right, the y values approach 3 so that .

169. 2

As x approaches –3 both from the left and from the right, the y values approach 2 so that .

Note that the actual value of f (–3) doesn't matter when you're finding the limit.

170. 3

As x approaches 1 from the left, the y values approach 3 so that .

171. 3

As x approaches 1 from the right, the y values approach 3 so that .

172. 5

As x approaches –2 both from the left and from the right, the y values approach 5 so that .

173. 4

Note that substituting the limiting value, 3, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and simplify:

Then substitute 3 for x:

174.

Note that substituting the limiting value, 2, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and denominator and simplify:

Then substitute 2 for x:

175.

Note that substituting the limiting value, –5, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and denominator and simplify:

Then substitute –5 for x:

176. 4

Note that substituting the limiting value, 4, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and simplify:

Then substitute 4 for x:

177. 1

Note that substituting in the limiting value, 0, gives you an indeterminate form. For example, as x approaches 0 from the right, you have the indeterminate form ∞ – ∞, and as x approaches 0 from the left, you have the indeterminate form –∞ + ∞.

To find the limit, first get common denominators and simplify:

Then substitute 0 for x:

178. 0

The given function is continuous everywhere, so you can simply substitute in the limiting value:

179. 3

Note that substituting the limiting value, –1, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and simplify:

Then substitute –1 for x:

180.

Note that substituting the limiting value, 0, into the function gives you the indeterminate form .

To find the limit, first multiply the numerator and denominator by the conjugate of the numerator and then simplify:

Then substitute 0 for h:

181. 108

Note that substituting the limiting value, 3, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and simplify:

Then substitute 3 for x:

182. ∞

Note that substituting in the limiting value gives you an indeterminate form.

Because x is approaching 0 from the right, you have x > 0 so that . Therefore, the limit becomes

As x → 0⁺, you have (2 + 2x) → 2 and x² → 0⁺ so that . The limit is positive infinity because dividing 2 by a very small positive number close to zero gives you a very large positive number.

183. ∞

Note that substituting in the limiting value gives you the indeterminate form ∞ – ∞.

Because x is approaching 0 from the left, you have x < 0 so that . Therefore, the limit becomes

Now consider the numerator and denominator as x → 0^–. As x → 0^–, you have (2 + 2x) → 2 and x² → 0⁺ so that . The limit is positive infinity because dividing 2 by a very small positive number close to zero gives you a very large positive number.

184. limit does not exist

Note that substituting in the limiting value gives you the indeterminate form .

Examine both the left-hand limit and right-hand limit to determine whether the limits are equal. To find the left-hand limit, consider values that are slightly smaller than and substitute into the limit. Notice that to simplify the absolute value in the denominator of the fraction, you replace the absolute values bars with parentheses and add a negative sign, because substituting in a value less than will make the number in the parentheses negative; the extra negative sign will make the value positive again.

You deal with the right-hand limit similarly. Here, when removing the absolute value bars, you simply replace them with parentheses; you don't need the negative sign because the value in the parentheses is positive when you're substituting in a value larger than .

Because the right-hand limit doesn't equal the left-hand limit, the limit doesn't exist.

185.

Note that substituting the limiting value, 5, into the function gives you the indeterminate form .

To find the limit, first factor the numerator and denominator and simplify:

Then substitute 5 for x:

186.

Note that substituting in the limiting value gives you the indeterminate form .

Begin by multiplying the numerator and denominator of the fraction by the conjugate of the numerator:

Next, factor –1 from the numerator and continue simplifying:

To find the limit, substitute 3 for x:

187. 12

Note that substituting the limiting value, 0, into the function gives you the indeterminate form .

Expand the numerator and simplify:

To find the limit, substitute 0 for h:

188. 1

Note that substituting in the limiting value, 4, gives you the indeterminate form .

Because x is approaching 4 from the right, you have x > 4 so that . Therefore, the limit becomes

189. –1

Note that substituting in the limiting value gives you the indeterminate form .

Because x is approaching 5 from the left, you have x < 5 so that . Therefore, the limit becomes

190.

Note that substituting in the limiting value, –5, into the function gives you the indeterminate form .

Begin by writing the two fractions in the numerator as a single fraction by getting common denominators. Then simplify:

To find the limit, substitute –5 for x:

191.

Note that substituting in the limiting value gives you an indeterminate form. For example, as x approaches 0 from the right, you have the indeterminate form ∞ – ∞, and as x approaches 0 from the left, you have the indeterminate form –∞ + ∞.

Begin by getting common denominators:

Next, multiply the numerator and denominator of the fraction by the conjugate of the numerator and simplify:

To find the limit, substitute 0 for x:

192.

Note that substituting in the limiting value, 0, gives you the indeterminate form .

Begin by rewriting the two terms in the numerator using positive exponents. After that, get common denominators in the numerator and simplify:

To find the limit, substitute 0 for h:

193. 5

The squeeze theorem states that if f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and if , then .

Note that and that . Therefore, by the squeeze theorem, .

194. 4

Note that and that . Therefore, by the squeeze theorem, .

195. 2

Note that and that . Therefore, by the squeeze theorem, .

196. 0

Notice that for all values of x except for x = 0, you have due to the range of the cosine function. So for all values of x except for x = 0, you have

Because x is approaching 0 (but isn't equal to 0), you can apply the squeeze theorem:

Therefore, you can conclude that .

197. 0

Notice that for all values of x > 0, you have due to the range of the sine function. So for all values of x > 0, you have

Because x is approaching 0 (but isn't equal to 0), you can apply the squeeze theorem:

Therefore, you can conclude that .

198. 0

Notice that for all values of x except for x = 0, you have due to the range of the sine function. So for all values of x except for x = 0, you have the following (after multiplying by –1):

You need to make the center expression match the given one, , so do a little algebra. Adding 3 gives you

Now multiply by :

Note that for values of x greater than 0, so you don't have to flip the inequalities.

Because the limit is approaching 0 from the right (but isn't equal to 0), you can apply the squeeze theorem to get

Therefore, you can conclude that .

199. 5

To use , you need the denominator of the function to match the argument of the sine. Begin by multiplying the numerator and denominator by 5. Then simplify:

200. 0

Factor the 2 from the numerator and then multiply the numerator and denominator by . Then simplify:

201.

Begin by using a trigonometric identity in the numerator and then factor:

Next, factor out a –1 from the numerator and simplify:

202.

You want to rewrite the expression so you can use . Begin by multiplying the numerator and denominator by and rewrite the expression as the product of two fractions:

Now you want a 5 in the denominator of the first fraction and a 9 in the numerator of the second fraction. Multiplying by and also by gives you

203.

Begin by rewriting tan(7x) to get

Now you want 7x in the denominator of the first fraction and 3x in the numerator of the third fraction. Therefore, multiply by , , and to get

204. 8

Begin by breaking up the fraction as

Next, you want each fraction to have a denominator of 2x, so multiply by , , and — or equivalently, by — to get

205.

Begin by factoring the denominator and rewriting the limit:

Notice that you can rewrite the first limit using the substitution so that as x → 2, you have ; this step isn't necessary, but it clarifies how to use in this problem.

Replacing (x – 2) with and replacing x → 2 with in the limit gives you the following:

206.

Begin by rewriting tan x and then multiply the numerator and denominator by . Then simplify:

Then substitute 0 into the equation and solve:

207. ∞

As x approaches 3 from the left, the y values approach ∞ so that .

208. –∞

As x approaches 3 from the right, the y values approach –∞ so that .

209. ∞

As x approaches 5 from the left, the y values approach ∞ so that .

210. –∞

As x approaches 5 from the right, the y values approach –∞ so that .

211. limit does not exist

As x approaches 5 from the left, the y values approach ∞. However, as x approaches 5 from the right, the y values approach –∞. Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

212. ∞

As x approaches 1 from the right, you have (x – 1) → 0⁺ so that

Note that the limit is positive infinity because dividing 3 by a small positive number close to zero gives you a large positive number.

213. –∞

As x approaches 1 from the left, you have (x – 1) → 0^– so that

Note that the limit is negative infinity because dividing 3 by a small negative number close to zero gives you a negative number whose absolute value is large.

214. –∞

Begin by writing the limit as . Then consider what happens in the numerator and the denominator as x approaches from the right. As , you have sin x → 1 and cos x → 0^–. Therefore, as , it follows that

215. ∞

Consider what happens in the numerator and denominator as x approaches π from the left. As x → π ^–, you have x² → π² and sin x → 0⁺. Therefore, as x → π ^–, it follows that

216. –∞

Consider what happens in the numerator and denominator as x approaches 5 from the left. As x → 5^–, you have (x + 3) → 8 and (x – 5) → 0^–. Therefore, as x → 5^–, it follows that

217. –∞

Consider what happens in the numerator and denominator as x approaches 0 from the left. As x → 0^–, you have (1 – x) → 1 and (e^x – 1) → 0^–. Therefore, as x → 0^–, it follows that

218. –∞

Begin by writing . Then consider what happens in the numerator and denominator as x approaches 0 from the left. As x → 0^–, you have cos x → 1 and sin x → 0^–. Therefore, as x → 0^–, it follows that

219. ∞

Consider what happens in the numerator and denominator as x approaches 2. As x → 2, you have 4e^x → 4e² and . Therefore, as x → 2, it follows that

220. –∞

Consider what happens to each factor in the numerator and denominator as x approaches from the right. As , you have and . Therefore, as , it follows that

221. –∞

Consider what happens in the numerator and denominator as x approaches 5. As x → 5, you have sin x → sin 5, and because π < 5 < 2π, it follows that sin 5 < 0. And as x → 5, you also have (5 – x)⁴ → 0⁺. Therefore, as x → 5, it follows that

Note that the limit is negative infinity because sin(5) is negative, and dividing sin(5) by a small positive number close to zero gives you a negative number whose absolute value is large.

222. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0. You need to examine both the left-hand limit and the right-hand limit.

For the left-hand limit, as x → 0^–, you have (x + 5) → 5, x⁴ → 0⁺, and (x – 6) → –6. Therefore, as x → 0^–, it follows that

For the right-hand limit, you have (x + 5) → 5, x⁴ → 0⁺, and (x – 6) → –6. Therefore, as x → 0⁺, it follows that

Because the left-hand limit is equal to the right-hand limit, .

223. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 1. Using the left-hand limit, as x → 1^–, you have (3x) → 3 and (e^x – e) → 0^–. Therefore, as x → 1^–, it follows that

Using the right-hand limit, as x → 1⁺, you have (3x) → 3 and (e^x – e) → 0⁺. So as x → 1⁺, it follows that

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

224. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0 from the right. You have (x – 1) → –1, x² → 0⁺, and (x + 2) → 2. Therefore, as x → 0⁺, you get the following:

225. –∞

Consider what happens in the numerator and denominator as x approaches e from the left. As x → e^–, you have x³ → e³ and (ln x – 1) → (ln e^– – 1) → 0^–. Therefore, as x → e^–, it follows that

226. limit does not exist

Begin by examining both the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches e². For the left-hand limit, as x → e^2–, you have –x → –e² and (ln x – 2) → 0^–. Therefore, as x → e^2–, it follows that

Note that the limit is positive infinity because dividing –e² by a small negative number close to zero gives you a large positive number.

For the right-hand limit, as x → e²⁺, you have –x → –e² and (ln x – 2) → 0⁺. Therefore, as x → e2⁺, it follows that

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

227. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 2. For the left-hand limit, as x → 2^–, you have (x + 2) → 4 and (x² – 4) → 0^–. Therefore, as x → 2^–, it follows that

For the right-hand limit, as x → 2⁺, you have (x + 2) → 4 and (x² – 4) → 0⁺. Therefore, as x → 2⁺, it follows that

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

228. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 25. For the left-hand limit, as x → 25^–, you have and (x – 25) → 0^–. Therefore, as x → 25^–, it follows that

For the right-hand limit, as x → 25⁺, you have and (x – 25) → 0⁺. Therefore, as x → 25⁺, you have

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

229. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0. As x → 0, you have (x² + 4) → 4, x² → 0⁺, and (x – 1) → –1. Therefore, as x → 0, it follows that

230. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0 from the left. You have (x – 1) → –1, x² → 0⁺, and (x + 2) → 2. Therefore, as x → 0^–, you get the following:

231. ∞

Consider what happens in the numerator and denominator as x approaches 3. As x → 3, you have (3 + x) → 6 and . Therefore, as x → 3, it follows that

232.

As the x values approach –∞, the y values approach so that .

233.

As the x values approach ∞, the y values approach so that .

234. 2

As the x values approach –∞, the y values approach 2 so that .

235. 2

As the x values approach ∞, the y values approach 2 so that .

236.

Divide the numerator and denominator by the highest power of x that appears in the denominator, x¹, and simplify:

Then apply the limit:

237. –∞

Begin by multiplying to get

For a polynomial function, the end behavior is determined by the term containing the largest power of x, so you have

238. 3

Divide the numerator and denominator by the highest power of x that appears in the denominator, x¹, and simplify:

Then apply the limit:

239. limit does not exist

Because cos x doesn't approach a specific value as x approaches ∞, the limit doesn't exist.

240. 0

Begin by expanding the denominator:

Next, divide the numerator and denominator by the highest power of x that appears in the denominator, x⁵, and simplify:

Then apply the limit:

241.

Begin by multiplying the numerator and denominator by so that you can simplify the expression underneath the square root:

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution in the denominator. Therefore, you have

Now apply the limit:

242. ∞

Divide the numerator and denominator by the highest power of x that appears in the denominator, x², and simplify:

Then apply the limit. Consider what happens in the numerator and the denominator asx → –∞. In the numerator, , and in the denominator, . Therefore, you have

243.

To simplify the expression underneath the radical, begin by multiplying the numerator and denominator by :

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution in the numerator:

Now apply the limit:

244.

To simplify the expression underneath the radical, begin by multiplying the numerator and denominator by :

Now apply the limit:

245. –∞

Notice that if you consider the limit immediately, you have the indeterminate form ∞ – ∞.

Begin by factoring to get

246.

Notice that if you consider the limit immediately, you have the indeterminate form ∞ – ∞.

Create a fraction and multiply the numerator and denominator by the conjugate of the expression . The conjugate is , so you have the following:

Next, multiply the numerator and denominator by so you can simplify the expression underneath the radical:

Now apply the limit:

247.

Notice that if you consider the limit immediately, you have the indeterminate form –∞ + ∞.

Create a fraction and multiply the numerator and denominator by the conjugate of the expression . The conjugate is , so you have

Next, multiply the numerator and denominator by so that you can simplify the expression underneath the square root:

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution in the denominator. Therefore, you have

Now apply the limit:

248.

To find any horizontal asymptotes of the function , you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by :

To find the limit as x → –∞, you can proceed in the same way in order to simplify:

Therefore, the only horizontal asymptote is .

249. y = –1

To find any horizontal asymptotes of the function , you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by :

To find the limit as x → –∞, proceed in the same way:

Therefore, the only horizontal asymptote is y = –1.

250.

In order to find any horizontal asymptotes of the function , you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by :

In order to find the limit as x → –∞, proceed in the same way, noting that as x → –∞, you still use :

Therefore, the only horizontal asymptote is .

251. y = 1 and y = –1

To find any horizontal asymptotes of the function , you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by :

To find the limit as x → –∞, proceed in the same way, noting that as , you need to use the substitution :

Therefore, the function has the horizontal asymptotes y = 1 and y = –1.

252. removable discontinuity at x = –3, jump discontinuity at x = 3

The limit exists at x = –3 but isn't equal to f (–3), which corresponds to a removable discontinuity.

At x = 3, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values); this corresponds to a jump discontinuity.

253. removable discontinuity at x = 1, infinite discontinuity at x = 5

The limit exists at x = 1, but f (1) is undefined, which corresponds to a removable discontinuity.

At x = 5, the left-hand limit is ∞ and the right-hand limit is –∞, so an infinite discontinuity exists at x = 5.

254. jump discontinuity at x = –2, jump discontinuity at x = 3

At x = –2, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values), which corresponds to a jump discontinuity.

At x = 3, the left-hand limit again doesn't equal the right-hand limit (both limits exist as finite values), so this also corresponds to a jump discontinuity.

255. removable discontinuity at x = –1, jump discontinuity at x = 4, infinite discontinuity at x = 6

At x = –1, the left-hand limit equals the right-hand limit, but the limit doesn't equal f (–1), which is undefined. Therefore, a removable discontinuity is at x = –1.

At x = 4, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values), so a jump discontinuity is at x = 4.

At x = 6, both the left and right-hand limits equal ∞, so an infinite discontinuity is at x = 6.

256. not continuous, infinite discontinuity

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is given by . As x → 2^–, you have (x – 2) → 0^– so that . Because the discontinuity is infinite, you don't need to examine the right-hand limit; you can conclude that the function is not continuous.

257. continuous, f (a) = 2

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is , and the right-hand limit at a is . The left-hand and right-hand limits match, so the limit at a exists and is equal to 2.

The value at a is . Because the function satisfies the definition of continuity, you can conclude that that function is continuous at a = 1.

258. continuous, f (a) = 5

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is

And the right-hand limit at a is

Note that in this case, you could have simply evaluated the limit as x approaches 3 instead of examining the left-hand limit and right-hand limit separately.

The left-hand and right-hand limits match, so the limit exists and is equal to 5. Because f (a) = f (3) = 5, the definition of continuity is satisfied and the function is continuous.

259. continuous,

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is

The right-hand limit at a is

The left-hand and right-hand limits match, so the limit exists.

Note that in this case, you could have simply evaluated the limit as x approaches 16 instead of examining the left-hand limit and right-hand limit separately.

The value at a is given by , which matches the limit. Because the definition of continuity is satisfied, you can conclude that that function is continuous at a = 16.

260. not continuous, jump discontinuity

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is

The right-hand limit at a is

Because the left- and right-hand limits exist but aren't equal to each other, there's a jump discontinuity at a = –6.

261. not continuous, removable discontinuity

A function f (x) is continuous at x = a if it satisfies the equation . The left-hand limit at a is

The right-hand limit at a is

The left-hand and right-hand limits match, so the limit exists and is equal to 3.

Note that in this case, you could have simply evaluated the limit as x approaches –1 instead of examining the left-hand limit and right-hand limit separately.

However, because f (a) = f (–1) = 2, the function is not continuous; it has a removable discontinuity.

262. continuous at a = 0 and at a = π

A function f (x) is continuous at x = a if it satisfies the equation . To determine whether the function is continuous at a = 0, see whether it satisfies the equation . The left-hand limit at a = 0 is , and the right-hand limit at a = 0 is . Because f (0) = 2 cos(0) = 2, the function is continuous at a = 0.

Likewise, decide whether the function satisfies the equation . The left-hand limit at a = π is , and the right-hand limit at a = π is . Because , the function is also continuous at a = π.

263. jump discontinuity at a = 1 and at a = 3

A function f (x) is continuous at x = a if it satisfies the equation . To determine whether the function is continuous at a = 1, see whether it satisfies the equation . The left-hand limit at a = 1 is , and the right-hand limit at a= 1 is . The limits differ, so there's a jump discontinuity.

Likewise, decide whether the function satisfies the equation . The left-hand limit at a = 3 is , and the right-hand limit at a = 3 is , so there's another jump discontinuity.

264. continuous at a = 2, jump discontinuity at a = 3

A function f (x) is continuous at x = a if it satisfies the equation . To determine whether the function is continuous at a = 2, see whether it satisfies the equation . The left-hand limit at a = 2 is , and the right-hand limit ata = 2 is . Because , the func-tion is continuous at .

Likewise, decide whether the function satisfies the equation . The left-hand limit at a = 3 is , and the right-hand limit at a = 3 is . The limits don't match, so there's a jump discontinuity at a = 3.

265. infinite discontinuity at a = 0, continuous at a = 4

A function f (x) is continuous at x = a if it satisfies the equation . To determine whether the function is continuous at a = 0, see whether it satisfies the equation . The left-hand limit at a = 0 is , and the right-hand limit at a= 0 is , so there's an infinite discontinuity at a = 0.

Likewise, decide whether the function satisfies the equation . The left-hand limit at a = 4 is , and the right-hand limit at a = 4 is . Because , the function is continuous at a = 4.

266.

To determine the value of c, you must satisfy the definition of a continuous function: .

The left-hand limit at x = 2 is given by , and the right-hand limit is given by . Also note that f (2) = 2c – 2.

The left-hand limit must equal the right-hand limit, so set them equal to each other and solve for c:

Therefore, is the solution.

267. c = 2

To determine the value of c, you must satisfy the definition of a continuous function: .

The left-hand limit at x = 4 is given by , and the right-hand limit is given by . Also note that .

The left-hand limit must equal the right-hand limit, so set them equal to each other:

Therefore, c – 2 = 0, which gives you the solution c = 2.

268. [1, 2]

Recall the intermediate value theorem: Suppose that f is continuous on the closed interval [a, b], and let N be any number between f (a) and f (b), where f (a) ≠ f (b). Then a number c exists in (a, b) such that f (c) = N.

Notice that is a polynomial that's continuous everywhere, so the interme-diate value theorem applies. Checking the endpoints of the interval [1, 2] gives you

Because the function changes signs on this interval, there's at least one root in the interval by the intermediate value theorem.

269. [16, 25]

Notice that is a continuous function for x ≥ 0, so the intermediate value theorem applies. Checking the endpoints of the interval [16, 25] gives you

Because the function changes signs on this interval, there's at least one root in the interval by the intermediate value theorem.

270. [2, 3]

The function f (x) = 2(3^x) + x² – 4 is continuous everywhere, so the intermediate value theorem applies. Checking the endpoints of the interval [2, 3] gives you

The number 32 is between 18 and 59, so by the intermediate value theorem, there exists at least one point c in the interval [2, 3] such that f (c) = 32.

271. [3, 4]

The function is continuous everywhere, so the intermediate value theorem applies. Checking the endpoints of the interval [3, 4] gives you

The number 22 is between 17 and 25, so by the intermediate value theorem, there exists at least one point c in the interval [3, 4] such that f (c) = 22.

272. x = 1

The graph has a point of discontinuity at x = 1, so the graph isn't differentiable there. However, the rest of the graph is smooth and continuous, so the derivative exists for all other points.

273. differentiable everywhere

Because the graph is continuous and smooth everywhere, the function is differentiable everywhere.

274. x = –2, x = 0, x = 2

The graph of the function has sharp corners at x = –2, x = 0, and x = 2, so the function isn't differentiable there.

You can also note that for each of the points x = –2, x = 0, and x = 2, the slopes of the tangent lines jump from 1 to –1 or from –1 to 1. This jump in slopes is another way to recognize values of x where the function is not differentiable.

275. , where n is an integer

Because y = tan x has points of discontinuity at , where n is an integer, the function isn't differentiable at those points.

276. x = 0

The tangent line would be vertical at x = 0, so the function isn't differentiable there.

277. 2

Use the derivative definition with f (x) = 2x – 1 and f (x + h) = 2(x + h) – 1 = 2x + 2h – 1:

278. 2 x

Use the derivative definition with f (x) = x² and f (x + h) = (x + h)² = (x + h)(x + h) = x² + 2xh + h²:

279. 1

Use the derivative definition with f (x) = 2 + x and f (x + h) = 2 + (x + h):

280.

Use the derivative definition with
and with
to get the following:

281. 3 x² – 2x

Using the derivative definition with f (x) = x³ – x² and with

gives you the following:

282. 6 x + 4

Using the derivative definition with f (x) = 3x² + 4x and with

gives you the following:

283.

Using the derivative definition with and with gives you the following:

284.

Using the derivative definition with and with gives you the following:

285.

Use the derivative definition with and with :

286. 3 x² + 3

Use the derivative definition with f (x) = x³ + 3x and with f (x + h) = (x + h)³ + 3(x + h) = x³ + 3x²h + 3xh² + h³ + 3x + 3h:

287.

Using the derivative definition with and with gives you the following:

288.

Using the derivative definition with and with gives you the following:

289.

Using the derivative definition with and with gives you the following:

290.

Using the derivative definition with and with gives you

To rationalize the numerator, consider the formula a³ – b³ = (a – b)(a² + ab + b²). If you let a = (2x + 2h + 1)^1/3 and b = (2x + 1)^1/3, you have (a – b) in the numerator; that means you can rationalize the numerator by multiplying by (a² + ab + b²):

291. 0

The tangent line at x = 3 is horizontal, so the slope is zero. Therefore, f'(3) = 0.

292. –1

The slope of the tangent line at x = –1 is equal to –1, so f'(–1) = –1.

293. 1

The slope of the tangent line at x = –3 is equal to 1, so f'(–3) =1.

294. 3

The slope of the tangent line at any point on the graph of y = 3x + 4 is equal to 3, so f'(–22π³) = 3.

295. f'(1) < f'(–2) < f'(–3)

The tangent line at x = –3 has a positive slope, the slope of the tangent line at x = –2 is equal to zero, and the slope of the tangent line at x = 1 is negative. Therefore, f'(1) < f'(–2) < f'(–3).

296. f'(1) < f'(2) < 0.1 < f'(5)

The slope of the tangent line at x = 1 is negative, the slope of the tangent line at x = 2 is equal to zero, and the slope of the tangent line at x = 5 is clearly larger than 0.1. Therefore, f'(1) < f'(2) < 0.1 < f'(5).

297. 5

Use basic derivative rules to get f'(x) = 5.

298. 2 x + 3

Apply the power rule to each term, recalling that the derivative of a constant is zero: f'(x) = 2x + 3.

299. 4 x + 7

Begin by multiplying the two factors together:

Then apply the power rule to get the derivative:

300. 0

Because π³ is constant, f'(x) = 0.

301.

Split up the radical and rewrite the power on the variable using exponential notation:

Then apply the power rule to find the derivative:

302.

Begin by breaking up the fraction:

Then apply the power rule to find the derivative:

303.

Begin by multiplying the factors together:

Then apply the power rule to find the derivative:

304.

Begin by rewriting the function using a negative exponent:

Then apply the power rule to each term to get the derivative:

305.

Apply the power rule to the first two terms of , recalling that the derivative of a constant is zero:

306.

Multiply the factors and rewrite the function using exponential notation:

Next, apply the power rule to find the derivative:

307.

Begin by multiplying the factors together:

Then find the derivative using the power rule:

308. 16 x³ – 2x + 8

Apply the power rule to each term, recalling that the derivative of a constant is zero: f'(x) = 16x³ – 2x + 8.

309.

Rewrite the function using exponential notation:

Then apply the power rule to each term to find the derivative:

310. , 1

Begin by finding the derivative of the function f (x) = x³ – x² – x + 1:

A horizontal tangent line has a slope of zero, so set the derivative equal to zero, factor, and solve for x:

Setting each factor equal to zero gives you 3x + 1 = 0, which has the solution , and gives you x – 1 = 0, which has the solution x = 1.

311.

Begin by finding the derivative of the function:

Next, set the derivative equal to 6 and solve for x:

312. 16 x⁷ + 5x⁴ – 8x³ – 1

Recall that the product rule states

You can multiply out the expression f (x) = (2x³ + 1)(x⁵ – 1) first and then avoid using the product rule, but here's how to find the derivative using the product rule:

313. x(2 sin x + x cos x)

Applying the product rule to f (x) = x² sin x gives you

314.

Apply the product rule to f (x) = sec x tan x:

315. (sec x)(1 + x tan x)

Begin by rewriting the original expression as and then apply the product rule:

316. 4 (csc x)(1 – x cot x)

Apply the product rule to f (x) = 4x csc x to get

317. 12

According to the product rule,

To find (fg)'(4), enter the numbers and solve:

318.

Recall that the product rule states

You can apply the quotient rule directly, or you can rewrite the original function as and then apply the product rule to get the following:

319. x sec x tan x + 2 sec³ x

Apply the product rule to f (x) = sec x(x + tan x) as follows:

320.

Apply the product rule to f (x) = (x² + x)csc x to get

321. 4 x² (sec x)(3 + x tan x)

Applying the product rule to f (x) = 4x³ sec x gives you

322.

You can apply the quotient rule directly, or you can rewrite the original function as and then apply the product rule as follows:

323.

Using the product rule on f (x) = x²g (x) and then factoring gives you the derivative as follows:

324.

Begin by breaking up the fraction and simplifying:

Next, apply the power rule to the first term and the product rule to the second term:

325. –46

The product rule tells you that

To find (fg)'(3), enter the numbers and solve:

326. 2 x cos x sin x – x² sin² x + x² cos² x

Recall that the product rule states

You can group the factors however you want and then apply the product rule within the product rule. If you group together the trigonometric functions and apply the product rule, you have f (x) = x²(cos x sin x) so that

327.

Begin by simplifying the given expression:

Then apply the product rule to the second term and the power rule to the first term:

328.

Rewrite the original expression:

Then apply the product rule to get the derivative:

329.

First simplify the given function:

Then apply the product rule to find the derivative:

330.

You can use the quotient rule directly, or you can rewrite the original expression as and then apply the product rule as follows:

331.

To find the derivative of , use the product rule within the product rule:

332.

Recall that the quotient rule states

Apply the quotient rule to :

333.

Apply the quotient rule to to get the derivative:

334.

Apply the quotient rule to :

335.

Apply the quotient rule to to get the following:

336.

The quotient rule gives you

To find , enter the numbers and solve:

337.

Recall that the quotient rule states

Apply the quotient rule to :

338.

Apply the quotient rule to to find the derivative:

339.

Apply the quotient rule to to get the derivative:

340.

First, simplify the numerator and denominator:

Then apply the quotient rule to get

341.

Apply the quotient rule to to get the following:

342.

Apply the quotient rule to :

343.

The quotient rule says that

To find , enter the numbers and solve:

344.

Recall that the quotient rule states

Apply the quotient rule to to get

345.

Apply the quotient rule to to get the following:

346.

Apply the quotient rule to :

347.

Apply the quotient rule to to get

348.

Apply the quotient rule to as follows:

Note that the identity sec² x – tan² x = 1 was used to simplify the final expression.

349.

First multiply the numerator and denominator by x:

Then use the quotient rule to find the derivative:

350.

Applying the quotient rule to , followed by factoring and simplifying, gives you the following:

351.

Applying the quotient rule to gives you

352.

Recall that the chain rule states

Applying the chain rule to gives you

353. 4 cos(4x)

Apply the chain rule to f (x) = sin(4x):

354.

Rewrite the function using exponential notation:

Then apply the chain rule:

355.

Rewrite the function as

Then apply the chain rule to get the derivative:

356.

First rewrite the function:

Then apply the chain rule:

357.

Apply the chain rule to to get

358.

Recall that the chain rule states

and that the quotient rule states

To find the derivative of , apply the quotient rule along with the chain rule:

359.

Recall that the chain rule states

and that the product rule states

To find the derivative of f (x) = cos(x sin x), apply the chain rule and the product rule:

360.

Rewrite the function using exponential notation:

Now apply the product rule, being careful to use the chain rule when taking the derivative of the second factor:

361.

Recall that the chain rule states

and that the quotient rule states

Rewrite the function using exponential notation:

Next, apply the quotient rule, making sure to use the chain rule when taking the derivative of the denominator:

362. 4 sec² x tan x

First rewrite the function:

Applying the chain rule gives you the derivative:

363.

Recall that the chain rule states

and that the quotient rule states

Rewrite the function using exponential notation:

Apply the quotient rule and the chain rule to get the derivative:

364.

Recall that the chain rule states

and that the product rule states

Rewrite the function using exponential notation:

Then apply the product rule and also use the chain rule on the second factor to get the derivative:

365.

To find the derivative of , use the product rule along with the chain rule and then factor:

366.

Recall that the chain rule states

and that the quotient rule states

Applying the quotient rule and the chain rule to gives you the following derivative:

367.

First rewrite the function using exponential notation:

Then apply both the chain rule and the quotient rule:

368.

To find the derivative of , apply the chain rule repeatedly:

369.

Rewrite the function using exponential notation:

Then apply the chain rule repeatedly to get the derivative:

370.

Recall that the chain rule states

and that the product rule states

Applying the product rule and chain rule to f (x) = (1 + 5x)⁴(2 + x – x²)⁷ and then factoring (factoring can be the tricky part!) gives you the following:

371. x = 0, π, 2π

Recall that the chain rule states

Begin by finding the derivative of the function f (x) = 2 cos x + sin² x:

Then set the derivative equal to zero to find the x values where the function has a horizontal tangent line:

Next, set each factor equal to zero and solve for x: sin x = 0 has the solutions x = 0, π, 2π, and cos x – 1 = 0, or cos x = 1, has the solutions x = 0, 2π. The slope of the tangent line is zero at each of these x values, so the solutions are x = 0, π, 2π.

372.

To find the derivative of , use the chain rule:

373. 28

Because the chain rule gives you , it follows that

374. –40

Because the chain rule gives you , it follows that

375.

Recall that the chain rule states

Using the chain rule on f (x) = H(x³) gives you the following:

Because , you know that that , so the derivative becomes

376. 80

Because the chain rule gives you , it follows that

377.

Use properties of logarithms to rewrite the function:

Then take the derivative:

378.

To find the derivative of f (x) = (ln x)⁴, apply the chain rule:

379.

Begin by using properties of logarithms to rewrite the function:

Then apply the chain rule to find the derivative:

380.

Begin by writing the function using exponential notation:

Then use the chain rule to find the derivative:

381.

To make the derivative easier to find, use properties of logarithms to break up the function:

Now apply the chain rule to each term to get the derivative:

382.

Begin by rewriting the function using properties of logarithms:

The derivative becomes

Note that log₇5 is a constant, so its derivative is equal to zero.

You can further simplify by using the change of base formula to write :

383. sec x

Applying the chain rule to gives you

384.

To find the derivative of , apply the quotient rule and chain rule to the first term and apply the chain rule to the second term:

385.

Begin by using properties of logarithms to rewrite the function:

Then apply the quotient rule and the chain rule:

You can now further simplify by using the change of base formula to write so that you have

386.

Begin by rewriting the function and taking the natural logarithm of each side:

Take the derivative of each side with respect to x:

Then multiply both sides by y:

Finally, replacing y with x^{tan x} gives you the answer:

387.

Begin by rewriting the function and taking the natural logarithm of each side:

Then take the derivative of each side with respect to x:

Multiplying both sides by y produces

And replacing y with gives you the answer:

388.

Begin by rewriting the function and taking the natural logarithm of each side:

Using properties of logarithms to expand gives you

Next, take the derivative of each side with respect to x:

Multiplying both sides by y produces the following:

Replacing y with and simplifying gives you the solution:

Note that you can find the derivative without logarithmic differentiation, but using it makes the math easier.

389.

Begin by rewriting the function and taking the natural logarithm of each side:

Use properties of logarithms to expand:

Then take the derivative of each side with respect to x:

Multiplying both sides by y produces

Replacing with y with gives you the answer:

Note that you can find the derivative without logarithmic differentiation, but using this technique makes the calculations easier.

390. 5 e^5x

Apply the chain rule to find the derivative of f (x) = e^5x:

391.

Applying the chain rule to gives you

392.

Applying the product rule to gives you the derivative as follows:

393. 2 x

Use properties of logarithms to simplify the function:

The derivative is simply equal to

394.

To find the derivative of , apply the product rule while also applying the chain rule to the first factor:

395.

Put the radical in exponential form:

Then apply the product rule as well as the chain rule to the first factor to get the derivative:

396.

Applying the chain rule to gives you the derivative as follows:

397.

Use properties of logarithms to break up the function:

Then apply the chain rule to each term to get the derivative:

398.

Apply the quotient rule to find the derivative of :

399.

First rewrite the function:

Applying the chain rule gives you the derivative as follows:

400.

To find the derivative of , apply the quotient rule while applying the chain rule to the numerator:

401.

To find the derivative of , apply the product rule while applying the chain rule to the second factor:

402. y = πx + 3

You can begin by finding the y value, because it isn't given:

Next, find the derivative of the function:

Substitute in the given x value to find the slope of the tangent line:

Now use the point-slope formula for a line to get the tangent line at x = 0:

403. y = x + 1

Begin by finding the derivative of the function f (x) = x² – x + 2:

Substitute in the given x value to find the slope of the tangent line:

Now use the point-slope formula for a line to get the tangent line at (1, 2):

404.

You can begin by finding the y value, because it isn't given:

Next, find the derivative of the function:

Substitute in the given x value to find the slope of the tangent line:

Now use the point-slope formula for a line to get the tangent line at x = 2:

405.

The normal line is perpendicular to the tangent line. Begin by finding the derivative of the function f (x) = 3x² + x – 2:

Then substitute in the given x value to find the slope of the tangent line:

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get .

Now use the point-slope formula for a line to get the normal line at (3, 28):

406.

The normal line is perpendicular to the tangent line. You can begin by finding the y value at , because it isn't given:

Next, find the derivative of the function:

Then substitute in the given x value to find the slope of the tangent line:

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get –1.

Now use the point-slope formula for a line to get the normal line at :

407.

The normal line is perpendicular to the tangent line. You can begin by finding the y value at x = e², because it isn't given:

Next, find the derivative of the function:

Then substitute in the given x value to find the slope of the tangent line:

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get .

Now use the point-slope formula for a line to get the normal line at x = e²:

408.

Taking the derivative of both sides of x² + y² = 9 and solving for gives you the following:

409.

Take the derivative of both sides of y⁵ + x²y³ = 2 + x²y and solve for :

410.

Take the derivative of both sides of x³y³ + x cos(y) = 7 and solve for :

411.

Take the derivative of both sides of and solve for :

412.

Take the derivative of both sides of and solve for :

413.

Take the derivative of both sides of and solve for :

414.

Begin by finding the first derivative of 8x² + y² = 8:

Next, find the second derivative:

Substituting in the value of gives you

And now using 8x² + y² = 8 gives you the answer:

415.

Begin by finding the first derivative of x⁵ + y⁵ = 1:

Next, find the second derivative:

Substituting in the value of gives you

And now using x⁵ + y⁵ = 1 gives you the answer:

416.

Begin by finding the first derivative of x³ + y³ = 5:

Next, find the second derivative:

Substituting in the value of gives you

And now using x³ + y³ = 5 gives you the answer:

417.

Begin by finding the first derivative of :

Next, find the second derivative:

Substituting in the value of gives you

And now using gives you the answer:

418. y = –x + 2

You know a point on the tangent line, so you just need to find the slope. Begin by finding the derivative of x² + xy + y² = 3:

Next, enter the values x = 1 and y = 1 and solve for the slope :

The tangent line has a slope of –1 and passes through (1, 1), so its equation is

419.

You know a point on the tangent line, so you just need to find the slope. Begin by finding the derivative of 3(x² + y²)² = 25(x² – y²):

Then enter the values x = 2 and y = 1 and solve for the slope :

The tangent line has a slope of and passes through (2, 1), so its equation is

420. y = –x + 1

You know a point on the tangent line, so you just need to find the slope. Begin by taking the derivative of x² + 2xy + y²:

Then use the values x = 0 and y = 1 and solve for the slope :

The tangent line has a slope of –1 and passes through (0, 1), so its equation is

421.

You know a point on the tangent line, so you just need to find the slope. Begin by taking the derivative of cos(xy) + x² = sin y:

Then use the values x = 1 and and solve for the slope :

The tangent line has a slope of and passes through , so its equation is

422. y = 1

You know a point on the tangent line, so you just need to find the slope. Begin by simplifying the left side of the equation.

Next, find the derivative:

Use the values x = 0 and y = 1 and solve for the slope :

The tangent line has a slope of 0 and passes through (0, 1), so its equation is

423.

Use the formula for the differential, dy = f '(x) dx:

So for the given values, x = 3 and , find dy:

424.

Use the formula for the differential, dy = f '(x) dx:

So for the given values, x = 1 and dx = –0.1, find dy:

425.

Use the formula for the differential, dy = f '(x) dx:

So for the given values, and dx = 0.02, find dy:

426. L(x) = 6x – 3

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with the slope m = 6 and the point (1, 3) gives you

Replace y with L(x) to get the solution: L(x) = 6x – 3.

427.

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with the slope m = –1 and the point gives you

Replace y with L(x) to get the solution: .

428.

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with the slope and the point (2, 6^1/3) gives you

Replace y with L(x) to get the solution:

.

429. 3.987

To estimate the value, you can find a linearization by using f (x) = x^2/3 and a = 8 and then substitute 7.96 into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with the slope and the point (8, 4) gives you

Replacing y with L(x) gives you the linearization

.

Finally, substitute in the value x = 7.96 to find the estimate:

430.

To estimate the value, you can find a linearization by using and a = 100 and then substitute 102 into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with slope and point (100, 10) gives you

Replacing y with L(x) gives you the linearization .

Finally, substitute in the value x = 102 to find the estimate:

431.

To estimate the value of tan 46°, you can find a linearization using f (x) = tan x and and then substitute the value into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

Next, find the derivative of the function to get the slope of the line:

Using the point-slope formula with slope m = 2 and point gives you

Replacing y with L(x) gives you the linearization .

Finally, substitute in the value of x = radians (that is, 46°) to find the estimate of tan 46°:

432.

The volume of a sphere is . To find the rate of expansion, take the derivative of each side with respect to time:

433. 8π m²/s

The area of a circle is A = πr². To find the rate of increase, take the derivative of each side with respect to time:

The circle increases at a rate of meter per second, so when radius r = 4 meters, the area increases at the following rate:

434. 508

Take the derivative of both sides with respect to t:

Then substitute in the given values, x = 3 and :

435.

Take the derivative of both sides with respect to t:

To find the value of z, enter x = 4 and y = 1 in the original equation:

Then substitute in the given values along with the value of z and solve for :

436. 2.49 m²/s

To find the area of the triangle, you need the base and the height. If you let the length of the base equal 8 meters, then one side of the triangle equals 6 meters. Therefore, if the height equals h, then so that .

Because the area of a triangle is , you can write the area as

This equation now involves , so you can take the derivative of both sides with respect to t to get the rate of increase:

Substitute in the given values, and radians/second, and simplify:

When the angle between the sides is , the area is increasing at a rate of about 2.49 square meters per second.

437. rad/s

The problem tells you how quickly the bottom of the ladder slides away from the wall , so first write an expression for x, the distance from the wall. Assuming that the ground and wall meet at a right angle, you can write

Taking the derivative with respect to time, t, gives you the rate at which the angle is changing:

Substitute in the given information, where and feet per second:

When the angle is , the angle is increasing at a rate of radians per second.

438. 72 cm²/min

The area of a triangle is , and the problem tells you how quickly the base and height are changing. To find the rate of the change in area, take the derivative of both sides of the equation with respect to time, t:

Note that you have to use the product rule to find the derivative of the right side of the equation.

Substitute in the given information, centimeters per minute, h = 32 centimeters, b = 20 centimeters, and centimeters per minute:

439. 13.56 km/h

Let x be the distance sailed by Ship A and let y be the distance sailed by Ship B. Using the Pythagorean theorem, the distance between the ships is

From the given information, you have kilometers per hour and kilometers per hour. Taking the derivative of both sides of the equation with respect to time gives you

Notice that after 3 hours have elapsed, x = 60 and y = 105. Therefore, using the Pythagorean theorem, you can deduce that kilometers.

Substitute in all these values and solve for :

At 3 p.m., the ships are moving apart at a rate of about 13.56 kilometers per hour.

440. 4.83 cm/s

From the Pythagorean theorem, the distance from the origin is D² = x² + y². Using the particle's path, , you get

Taking the derivative of both sides with respect to time gives you

From the given values x = 8 centimeters and y = 3 centimeters, you find that the distance from the origin is centimeters. Use these values along with centimeters per second and solve for :

441. 2.95 mi/h

Let x be the distance that the person who is walking west has traveled, let y be the distance that the person traveling southwest has traveled, and let z be the distance between them.

Use the law of cosines to relate x, y, and z:

Take the derivative of both sides of the equation with respect to time:

After 40 minutes, or hour, and . Using the initial equation, you find that z ≈ 1.96 miles at this time. Substitute these values into the derivative and solve for :

After 40 minutes, the distance between these people is increasing at a rate of about 2.95 miles per hour.

442.

The water in the trough has a volume of . As the water level rises, the base and height of the triangle shapes increase. By similar triangles, you have so that . Therefore, the volume is

Taking the derivative with respect to time gives you

Use h = 1 foot and cubic feet per minute to find how quickly the water level is rising:

When the water is 1 foot deep, the water level is rising at a rate of feet per minute.

443. 698.86 km/h

Let x be the distance that the jet travels, and let y be the distance between the jet and the radar station.

Use the law of cosines to relate x and y:

Taking the derivative of both sides of the equation with respect to time gives you

After 2 minutes, the jet has traveled kilometers, so the distance from the radar station is

Enter the value of y in the equation and solve:

After 2 minutes, the jet is moving away from the radar station at a rate of about 698.86 kilometers per hour.

444.

Let x be the distance from the Point P to the spot on the shore where the light is shining. From the diagram, you have , or . Taking the derivative of both sides of the equation with respect to time gives you

When x = 2, . Using a trigonometric identity, you have

From the given information, you know that revolutions per minute. Because 2π radians are in one revolution, you can convert as follows: radians per minute:

445.

The volume of a cone is . The problem tells you that cubic feet per minute and that d = 2h. Because the diameter is twice the radius, you have 2r = d = 2h so that r = h; therefore, the volume becomes

Take the derivative with respect to time:

Substitute in the given information and solve for :

When the pile is 12 feet high, the height of the pile is increasing at a rate of (about 0.04) feet per minute.

446. no absolute maximum; absolute minimum: y = –1; no local maxima; local minimum: (1, –1)

There's no absolute maximum because the graph doesn't attain a largest y value. There are also no local maxima. The absolute minimum is y = –1. The point (1, –1) is a local minimum.

447. absolute maximum: y = 4; absolute minimum: y = 0; local maximum: (5, 4); local minima: (1, 0), (7, 2)

The absolute maximum value is 4, which the graph attains at the point (5, 4). The absolute minimum is 0, which is attained at the point (1, 0). The point (5, 4) also corresponds to a local maximum, and the local minima occur at (1, 0) and (7, 2).

448. absolute maximum: y = 3; no absolute minimum; local maxima: (3, 3), (5, 3); local minimum: (4, 1)

The absolute maximum value is 3, which the graph attains at the points (3, 3) and (5, 3). The function approaches the x-axis but doesn't cross or touch it, so there's no absolute minimum. The points (3, 3) and (5, 3) also correspond to local maxima, and the local minimum occurs at (4, 1).

449. no maxima or minima

The graph has no absolute maximum or minimum because the graph approaches ∞ on the left and –∞ on the right. Likewise, there are no local maxima or minima because (2, 1) and (5, 3) are points of discontinuity.

450. I and II

Because Point A satisfies the definition of both a local maximum and a local minimum, it's both. The graph decreases to negative infinity on the left side, so Point A is not an absolute minimum.

451. absolute maximum: 5; absolute minimum: –7

Begin by finding the derivative of the function; then find any critical numbers on the given interval by determining where the derivative equals zero or is undefined. Note that by finding critical numbers, you're finding potential turning points or cusp points of the graph.

The derivative of f (x) = 3x² – 12x + 5 is

Setting the derivative equal to zero and solving gives you the only critical number, x = 2. Next, substitute the endpoints of the interval and the critical number into the original function and pick the largest and smallest values:

Therefore, the absolute maximum is 5, and the absolute minimum is –7.

452. absolute maximum: 67; absolute minimum: 3

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of f (x) = x⁴ – 2x² + 4 is

Setting the derivative equal to zero and solving gives you the critical numbers x = 0 and x = ±1, all of which fall within the given interval. Next, substitute the endpoints of the interval along with the critical numbers into the original function and pick the largest and smallest values:

Therefore, the absolute maximum is 67, and the absolute minimum is 3.

453. absolute maximum: ; absolute minimum: 0

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of is

Next, find the critical numbers by setting the numerator equal to zero and solving for x (note that the denominator will never be zero):

Only x = 1 is in the given interval, so don't use x = –1. Next, substitute the endpoints of the interval along with the critical number into the original function and pick the largest and smallest values:

Therefore, the absolute maximum is , and the absolute minimum is 0.

454. absolute maximum: 2; absolute minimum:

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of is

Next, find any critical numbers of the function by setting the numerator and denominator of the derivative equal to zero and solving for t.

From the numerator of the derivative, you have 4 – 2t² = 0 so that 4 = 2t², or 2 = t², which has the solutions . From the denominator, you have 4 – t² = 0, or 4 = t², which has the solutions t = ±2.

Substitute the endpoints along with the critical points that fall within the interval into the original function and pick the largest and smallest values:

Therefore, the absolute maximum is 2, and the absolute minimum is .

455. absolute maximum: π + 2; absolute minimum:

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of f (x) = x – 2 cos x is

Setting the derivative equal to zero in order to find the critical numbers gives you 1+ 2 sin x = 0, or , which has the solutions and in the given interval of [–π, π].

Substitute these critical numbers along with the endpoints of the interval into the original function and pick the largest and smallest values:

Therefore, the absolute maximum is π + 2, and the absolute minimum is .

456. increasing on (–∞, –2) and (2, ∞); decreasing on (–2, 2)

Begin by finding the derivative of the function; then find any critical numbers on the given interval by determining where the derivative equals zero or is undefined. Note that by finding critical numbers, you're finding potential turning points or cusp points of the graph.

The derivative of the function f (x) = 2x³ – 24x + 1 is

Setting the derivative equal to zero in order to find the critical points gives you x² – 4 = 0, or x² = 4, so that x = ±2.

To determine where the function is increasing or decreasing, substitute a test point inside each interval into the derivative to see whether the derivative is positive or negative.

To test (–∞, –2), you can use x = –3. In that case, f '(–3) = 6(–3)² – 24 = 30. The derivative is positive, so the function is increasing on (–∞, –2). Proceed in a similar manner for the intervals (–2, 2) and (2, ∞). When x = 0, you have f '(0) = 6(0)² – 24 = –24, so the function is decreasing on (–2, 2). And if x = 3, then f '(3) = 6(3)² – 24 = 30, so the function is increasing on (2, ∞).

457. increasing on (–2, ∞); decreasing on (–3, –2)

Begin by finding the derivative of the function. The derivative of is

Next, find any critical numbers that fall inside the interval (–3, ∞) by setting the derivative equal to zero and solving for x. (You only need to set the numerator equal to zero, because the denominator will never equal zero.) This gives you 3x + 6 = 0, or x = –2.

Next, pick a value in (–3, –2) and determine whether the derivative is positive or negative. So if x = –2.5, then ; therefore, the function is decreasing on (–3, –2). Likewise, you can show that f '(x) > 0 on (–2, ∞), which means that f (x) is increasing on (–2, ∞).

458. increasing on , ; decreasing on , ,

Begin by finding the derivative of the function. The derivative of f (x) = cos² x – sin x is

Next, find any critical numbers that fall inside the specified interval by setting the derivative equal to zero and solving: –2 cos x sin x = 0, or (–cos x)(2 sin x + 1) = 0. Solving –cos x = 0 gives you the solutions and , and solving 2 sin + 1 = 0 gives you , which has the solutions and .

To determine where the original function is increasing or decreasing, substitute a test point from inside each interval into the derivative to see whether the derivative is positive or negative. So for the interval , you can use . In that case, , so the function is decreasing on . Proceeding in a similar manner, you can show that for a point inside the interval , f ' > 0; that for a point inside the interval , f '> 0; that for a point inside , f ' > 0; and that for a point inside , f ' > 0. Therefore, f (x) is increasing on and , and f (x) is decreasing on , , and .

459. increasing on , ; decreasing on ,

Begin by finding the derivative of the function. The derivative of f (x) = 2 cos x – cos 2x is

Now find the critical numbers by setting each factor equal to zero and solving for x. The equation sin x = 0 gives you the solutions x = 0, x = π, and x = 2π. Solving 2 cos x – 1 = 0 gives you , which has solutions and .

To determine whether the function is increasing or decreasing on , pick a point in the interval and substitute it into the derivative. So if you use , then you have , so the function is increasing on . Likewise, if you use the point in the interval , you have . If you use in the interval , then . And if you use in the interval , you have . Therefore, f (x) is increasing on the intervals and , and f (x) is decreasing on the intervals and .

460. increasing on (0, 1); decreasing on (1, ∞)

Begin by finding the derivative of the function. The derivative of f (x) = 4 ln x – 2x² is

Now find the critical numbers. Setting the numerator equal to zero gives you (1 – x)(1 + x) = 0 so that x = 1 or –1. Setting the denominator of the derivative equal to zero gives you x = 0. Notice that neither x = 0 nor x = –1 is in the domain of the original function.

To determine whether the function is increasing or decreasing on (0, 1), take a point in the interval and substitute it into the derivative. If , then , so the function is increasing on (0, 1). Likewise, if you use x = 2, then , so the function is decreasing on (1, ∞).

461. local maximum at (–1, 7); local minimum at (2, –20)

Begin by finding the derivative of the function f (x) = 2x³ – 3x² – 12x:

Then set this derivative equal to zero and solve for x to find the critical numbers:

Next, determine whether the function is increasing or decreasing on the intervals (–∞, –1), (–1, 2), and (2, ∞) by picking a point inside each interval and substituting it into the derivative. Using the values x = –2, x = 0, and x = 3 gives you

Therefore, f (x) is increasing on (–∞, –1), decreasing on (–1, 2), and increasing again on (2, ∞). That means there's a local maximum at x = –1 and a local minimum at x = 2.

Now enter these values in the original function to find the coordinates of the local maximum and minimum. Because f (–1) = 2(–1)³ – 3(–1)² – 12(–1) = 7, the local maximum occurs at (–1, 7). Because f (2) = 2(2)³ – 3(2)² – 12(2) = –20, the local minimum occurs at (2, –20).

462. no local maxima; local minimum at (16, –16)

Begin by finding the derivative of the function :

Then find the critical numbers. Setting the numerator equal to zero gives you so that , or x = 16. Setting the denominator equal to zero gives you x = 0. Next, determine whether the function is increasing or decreasing on the intervals (0, 16) and (16, ∞) by taking a point in each interval and substituting it into the derivative. So if x = 1, you have so that f (x) is decreasing on (0, 16). And if x = 25, you have , so f (x) is increasing on (16, ∞). Therefore, the function has a local minimum at x = 16. Finish by finding the coordinates: , so the local minimum is at (16, –16).

463. local maximum at (64, 32); local minimum at (0, 0)

Begin by finding the derivative of the function f (x) = 6x^2/3 – x:

Then find the critical numbers. Setting the numerator equal to zero gives you so that 4 = x^1/3, or 64 = x. Setting the denominator equal to zero gives you x = 0 as a solution.

Next, determine whether the function is increasing or decreasing on the intervals (–∞, 0), (0, 64), and (64, ∞) by taking a point in each interval and substituting it into the derivative. Using the test points x = –1, x = 1, and x = 125 gives you the following:

Therefore, f (x) is decreasing on (–∞, 0), increasing on (0, 64), and decreasing on (64, ∞). That means f (x) has a local minimum at x = 0; f (0) = 0, so a local minimum occurs at (0, 0). Also, f (x) has a local maximum at x = 64; f (64) = 6(64)^2/3 – 64 = 32, so the local maximum occurs at (64, 32).

464. local maximum at ; local minimum at

Begin by finding the derivative of the function f (x) = 2 sin x – sin 2x:

Next, find the critical numbers by setting the derivative equal to zero:

Setting the first factor equal to zero gives you , which has the solutions and . Setting the second factor equal to zero gives you , which has the solutions x = 0 and 2π.

Next, determine whether the function is increasing or decreasing on the intervals , , and by taking a point inside each interval and substituting it into the derivative to see whether it's positive or negative. Using the points , π, and gives you

So f (x) has a local maximum when because the function changes from increasing to decreasing at this value, and f (x) has a local minimum when because the function changes from decreasing to increasing at this value.

To find the points on the original function, substitute in these values. If , then

So the local maximum occurs at . If , then

Therefore, the local minimum occurs at .

465. local maxima at , ; local minima at ,

Begin by finding the derivative of the function f (x) = x + 2 cos x:

Next, find the critical numbers by setting the function equal to zero and solving for x on the given interval:

Then determine whether the function is increasing or decreasing on the intervals , , , , and by taking a point inside each interval and substituting it into the derivative. Using the value x = –6.27, which is slightly larger than –2π, gives you f '(–6.27) ≈ f '(–2π) = 1 > 0. Likewise, using the points , 0, , and from each of the remaining four intervals gives you the following:

Therefore, the function has local maxima when and and local minima when and . To find the points on the original function, substitute these x values into the original function:

Therefore, the local maxima are and , and the local minima are and .

466. concave up on (1, ∞); concave down on (–∞, 1)

To determine concavity, examine the second derivative. Because a derivative measures a “rate of change” and the first derivative of a function gives the slopes of tangent lines, the derivative of the derivative (the second derivative) measures the rate of change of the slopes of the tangent lines. If the second derivative is positive on an interval, the slopes of the tangent lines are increasing, so the function is bending upward, or is concave up. Likewise, if the second derivative is negative on an interval, the slopes of the tangent lines are decreasing, so the function is bending downward, or is concave down.

Begin by finding the first and second derivatives of the function f (x) = x³ – 3x² + 4:

Setting the second derivative equal to zero gives you 6x – 6 = 0 so that x = 1.

To determine the concavity on the intervals (–∞, 1) and (1, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = 0 and x = 2, you have f "(0) = 6(0) – 6 > 0 and f "(2) = 6(2) – 6 > 0. Therefore, f (x) is concave up on the interval (1, ∞) and concave down on the interval (–∞, 1).

467. concave up nowhere; concave down on (–∞, 0), (0, ∞)

Begin by finding the first and second derivatives of the function f (x) = 9x^2/3 – x:

Next, set the denominator of the second derivative equal to zero to get x^4/3 = 0 so that x = 0.

To determine the concavity on the intervals (–∞, 0) and (0, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1 and x = 1, you have the following:

Therefore, the f (x) is concave down on the intervals (–∞, 0) and (0, ∞).

468. concave up on (–∞, 0), ; concave down on

Begin by finding the first derivative of the function f (x) = x^1/3(x + 1) = x^4/3 + x^1/3:

Next, use the power rule to find the second derivative:

Then find where the second derivative is equal to zero or undefined. Setting the numerator equal to zero gives you 2x – 1 = 0 so that , and setting the denominator equal to zero gives you 9x^5/3 = 0 so that x = 0.

To determine the concavity on the intervals (–∞, 0), , and , pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1, , and x = 1 gives you the following:

Therefore, f (x) is concave up on the intervals (–∞, 0) and , and f (x) is concave down on the interval .

469. concave up on (–∞, –2), , (2, ∞); concave down on ,

Begin by finding the first and second derivatives of the function f (x) = (x² – 4)³:

Next, set each factor in the second derivative equal to zero and solve for x: x² – 4 = 0 gives you the solutions x = 2 and x = –2, and 5x² – 4 = 0, or , has the solutions and .

To determine the concavity on the intervals (–∞, –2), , , , and (2, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –3, x = –1, x = 0, x = 1, and x = 3, you have the following:

Therefore, f (x) is concave up on the intervals (–∞, –2), , and (2, ∞), and f (x) is concave down on the intervals and .

470. concave up on , ; concave down on the intervals (0, 0.253), ,

Begin by finding the first and second derivatives of the function f (x) = 2 cos x – sin(2x):

Next, set each factor of the second derivative equal to zero and solve: cos x = 0 has the solutions and , and 4 sin x – 1 = 0, or , has the solutions and x = π – 0.253.

To determine the concavity on the intervals (0, 0.25), , , , and , pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = 0.1, , , x = π, and gives you the following:

Therefore, f (x) is concave up on the intervals and , and f (x) is concave down on the intervals (0, 0.253), , and .

471. no inflection points

Inflection points are points where the function changes concavity. To determine concavity, examine the second derivative. Because a derivative measures a “rate of change” and the first derivative of a function gives the slopes of tangent lines, the derivative of the derivative (the second derivative) measures the rate of change of the slopes of the tangent lines. If the second derivative is positive on an interval, the slopes of the tangent lines are increasing, so the function is bending upward, or is concave up. Likewise, if the second derivative is negative on an interval, the slopes of the tangent lines are decreasing, so the function is bending downward, or is concave down.

Begin by finding the first and second derivatives of the function . The first derivative is

Now find the second derivative:

Notice that the numerator of the second derivative never equals zero and that the denominator equals zero when x = ±3; however, x = ±3 isn't in the domain of the original function. Therefore, there can be no inflection points.

472.

Begin by finding the first and second derivatives of the function f (x) = 2x³ + x². The first derivative is

And the second derivative is

Setting the second derivative equal to zero gives you 12x + 2 = 0, which has the solution .

To determine the concavity on the intervals and , pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1 and x = 0 gives you the following:

Therefore, the function is concave down on the interval and concave up on the interval , so is an inflection point. The corresponding y value on f (x) is

Therefore, the inflection point is .

473. no inflection points

Being by finding the first and second derivatives of the function . The first derivative is

And the second derivative is

Next, find where the second derivative is equal to zero or undefined. Setting the numerator equal to zero gives you sin x = 0, which has solutions x = 0, x = π, and x = 2π. Setting the denominator equal to zero gives you 1 + cos x = 0, or cos x = –1, which has the solution x = π.

To determine the concavity on the intervals (0, π) and (π, 2π), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values and gives you the following:

However, note that at x = π, the original function is undefined, so there are no inflection points. In fact, if you noticed this at the beginning, there's really no need to determine the concavity on the intervals!

474. (π, 0)

Begin by finding the first and second derivatives of the function f (x) = 3 sin x – sin³ x. The first derivative is

And the second derivative is

Next, find where the second derivative is equal to zero by setting each factor equal to zero: cos x has the solutions and , and sin x = 0 has the solutions x = 0, x = π, and x = 2π.

To determine the concavity on the intervals , , , and , pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values , , , and , you have the following:

Therefore, the concavity changes when x = π. The corresponding y value on f (x) is f (π) = 3 sin π – (sin π)³ = 0, so the inflection point is (π, 0).

475. (–1, –6)

Begin by finding the first and second derivatives of the function f (x) = x^5/3 – 5x^2/3. The first derivative is

And the second derivative is

Next, find where the second derivative is equal to zero or undefined by setting the numerator and the denominator equal to zero. For the numerator, you have 10x + 10 = 0, which has the solution x = –1. For the denominator, you have 9x^4/3 = 0, which has the solution x = 0.

To determine the concavity on the intervals (–∞, –1), (1, 0), and (0, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –2, , and x = 1 gives you

Because the concavity changes at x = –1, the original function has an inflection point there. The corresponding y value on f (x) is

Therefore, the inflection point is (–1, –6).

476. no local maxima; local minimum at (0, 1)

Begin by finding the first derivative of the function :

Next, find any critical numbers of the function. Setting the numerator of the derivative equal to zero gives you the solution x = 0. Note that no real values can make the denominator equal to zero.

Then find the second derivative:

To see whether the original function is concave up or concave down at the critical number, substitute x = 0 into the second derivative:

The second derivative is positive, so the original function is concave up at the critical point; therefore, a local minimum is at x = 0. The corresponding y value on the original function is , so the local minimum is at (0, 1).

477. local maximum at (0, 1); local minima at ,

Begin by finding the first derivative of the function f (x) = x⁴ – 4x² + 1:

Set the first derivative equal to zero and solve for x to find the critical numbers of f. The critical numbers are x = 0, , and .

Next, find the second derivative:

Substitute each of the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

Therefore, the original function has a local maximum when x = 0 and local minima when and . The corresponding y values on the original function are , f (0) = 0⁴ – 4(0)² + 1 = 1, and . Therefore, the local maximum occurs at (0, 1), and the local minima occur at and .

478. local maxima at , ; local minimum at (0, 0)

Begin by finding the first derivative of the function f (x) = 2x²(1 – x²) = 2x² – 2x⁴:

Set the first derivative equal to zero and solve for x to find the critical numbers. Solving 4x(1 – 2x²) = 0 gives you x = 0, , and as the critical numbers.

Next, find the second derivative:

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

So the original function has local maxima when and and a local minimum when x = 0. Finding the corresponding y values on the original function gives you the following:

Therefore, the local maxima are at and , and the local minimum is at (0, 0).

479. local maximum at ; local minimum at

Begin by finding the first derivative of the function :

Then set the numerator and denominator of the first derivative equal to zero to find the critical numbers. Setting the numerator equal to zero gives you –x² + 4 = 0, which has the solutions x = 2 and x = –2. Notice that the denominator of the derivative doesn't equal zero for any value of x.

Next, find the second derivative of the function:

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

Because f (x) is concave up when x = –2 and concave down when x = 2, a local minimum occurs at x = –2, and a local maximum occurs when x = 2. Because and , the local minimum occurs at , and the local maximum occurs at .

480. local maximum at ; local minimum at

Begin by finding the first derivative of the function f (x) = 2 sin x – x:

Set the first derivative equal to zero to find the critical numbers. From the equation 2 cos x – 1 = 0, you get , so the critical numbers are and .

Next, find the second derivative:

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at these values:

Because the function is concave down at , it has a local maximum there, and because the function is concave up at , it has a local minimum there. To find the corresponding y value on f (x), substitute the critical numbers into the original function:

Therefore, the local maximum is at , and the local minimum is at .

481.

Recall Rolle's theorem: If f is a function that satisfies the following three hypotheses:

· f is continuous on the closed interval [a, b]

· f is differentiable on the open interval (a, b)

· f (a) = f (b)

then there is a number c in (a, b) such that f '(c) = 0.

Notice that the given function is differentiable everywhere (and therefore continuous everywhere) because it's a polynomial. Then verify that f (0) = f (6) so that Rolle's theorem can be applied:

Next, find the derivative of the function, set it equal to zero, and solve for c:

482.

Notice that the given function is differentiable on (–8, 0). Then verify that f (–8) = f (0) so that Rolle's theorem can be applied:

Next, find the derivative of the function, set it equal to zero, and solve. Here's the derivative:

Setting the numerator equal to zero and solving for c gives you

483. 0, , ±1

Notice that the given function is differentiable everywhere. Then verify that f (–1) = f (1) so that Rolle's theorem can be applied:

Next, find the derivative of the function, set it equal to zero, and solve for c. Here's the derivative:

You need to find the solutions of sin(2πc) = 0 on the interval –1 ≤ c ≤ 1 so that –2π ≤ 2πc ≤ 2π. Notice that the solutions occur when 2πc = –2π, –π, 0, π, and 2π. Solving each of these equations for c gives you the solutions c = –1, , c = 0, , and c = 1.

484.

Recall the mean value theorem: If f is a function that satisfies the following hypotheses:

· f is continuous on the closed interval [a, b]

· f is differentiable on the open interval (a, b)

then there is a number c in (a, b) such that .

Notice that the given function is differentiable everywhere, so you can apply the mean value theorem. Next, find the value of on the given interval:

Now find the derivative of f (x) = x³ + 3x – 1, the given function, and set it equal to 7. The derivative is f ' = 3x² + 3, so

The negative root falls outside the given interval, so keeping only the positive root gives you the solution .

485.

Notice that the given function is differentiable on (0, 1) and continuous on [0, 1], so you can apply the mean value theorem. Next, find the value of on the given interval:

Now find the derivative of , the given function, and set it equal to 2. The derivative is , so you have

486.

Notice that the given function is differentiable everywhere, so you can apply the mean value theorem. Next, find the value of on the given interval:

Now find the derivative of , the given function, and set it equal to . The derivative is , so you have

Keeping only the positive root, you have , which has the solution . Note that the negative root would give you , which is outside of the interval.

487. 24

By the mean value theorem, you have the following equation for some c on the interval [1, 5]:

Next, solve for f (5) using the fact that f (1) = 12 and f '(x) ≥ 3 (and so f '(c) ≥ 3):

488. 8 ≤ f (8) – f (4) ≤ 24

If 2 ≤ f '(x) ≤ 6, then by the mean value theorem, you have the following equation for some c in the interval [4, 8]:

Using 2 ≤ f '(x) ≤ 6, you can bound the value of 4f '(c) because 2 ≤ f '(c) ≤ 6:

Because f (8) – f (4) = 4f '(c), it follows that

489.

The given function is continuous and differentiable on the given interval. By the mean value theorem, there exists a number c in the interval [8, 9] such that

But because , you have . Replacing in the equation gives you

Because , you have , so

Notice also that f ' is decreasing (f " will be negative on the given interval), so the following is true:

Because , you have . Therefore, it follows that

490. velocity: 2; acceleration: 2

Begin by taking the derivative of the position function s(t) to find the velocity function:

Substituting in the value of t = 5, find the velocity:

Next, take the derivative of the velocity function to find the acceleration function:

Because the acceleration is a constant at t = 5, the acceleration is a(5) = 2.

491. velocity: 1; acceleration: –2