1. C Use the double angle formula for sine, sin 2θ = 2 sinθ cosθ. to rewrite the limit and then solve:

2. A If the highest power of x in a rational expression is the same in both the numerator and the denominator then the limit as x approaches infinity is the coefficient of the highest term in the numerator divided by the coefficient of the highest term in the denominator. The coefficient of the highest term in the numerator is 1 and the coefficient of the highest term in the denominator is 3, so the limit is .

3. B A removable discontinuity occurs when a rational expression has common factors in the numerator and the denominator. The reduced function has the factor (x + 2) in the numerator and denominator, hence there is a removable discontinuity when x = –2. The y-coordinate of the discontinuity is found by plugging x = –2 into the reduced function, f(x) = . Thus, the point where a removable discontinuity exists is (–2,–1).

4. D There are three conditions a function must fulfill for it to be continuous at a point x = c: 1) f(c) exists, 2) f(x) exists, and 3) f(x) = f(c). All of the functions above satisfy all three conditions, except j(x). j(–3) does not exist (and the other two conditions are not met either).

5. E There are three conditions a function must fulfill for it to be continuous at a point x = c: f(c) exists, f(x) = exists, and f(x) = f(c). Answer choice E is the only one that satisfies all three. F(c) does not exist in answer choice D. f(x) does not exist in answer choices A or C. f(x) ≠ f(c) in answer choice B.

6. E Recall the definition of the derivative says: = f′(x) and the derivative of sec x is tan x sec x. Thus, = = = . Therefore, the limit does not exist.

7. A Using the Product Rule, , take the derivative of f (x) and you get A.

8. E Using the Chain Rule, , take the derivative of f(x) and you get E.

9. C First, use implicit differentiation to find :

2x²y + 2xy² = 0

Isolate and simplify:

Next, take the second derivative via implicit differentiation:

Plug in :

There is no need to simplify, just plug in the point (2,1):

10. B Start by plugging (0,2) into y = ax² + bx + c. Then, c = 2. Next, take the derivative of y = ax² + bx + c: = 2ax + b. Given that y = x + 2 is normal to y = ax² + bx + c, the slope of y = ax² + bx + c, or , is –3. Thus, by plugging x = 0 into = 2ax + b, b = –3. Finally, plug the point (2,1) and the values of c and b into y = ax² + bx + c. Therefore, a = .

11. A Plug h(x) in as x into g(x) and take the derivative using the Chain Rule, so (2g(x³)) = 6x²g(x³).

12. D In order to determine the number of relative extrema, the derivative of f(x) must be set equal to zero and the critical points found. When this is done, f′(x) = 6x⁴ − 6x² and the critical points are located at x = –1, x = 0, and x = 1. Therefore, there are three possible relative extrema for this curve, and answer choices A and C can be eliminated. Next, in order to determine the points of inflection, set the second derivative of f(x) equal to zero and solve. When this is done, f″(x) = 24x³ − 12x and the points of inflection are located at x = 0, x = − and x = . Thus, there are three points of inflection and the answer is D.

13. A Use u-substitution. Here, u = 2x – 5 and du = 2 dx. Then, . Replace u for the final solution: + C.

14. B Use u-substitution. Here u = x³ − 6, and du = 4x² dx. Then, . Replace u for the final solution: + C.

15. A Use the mean value theorem for integrals, f(c) = . Thus, for this problem, .

16. E Use the Second Fundamental Theorem of Calculus: . Thus, for this problem, = 6x((3x²)² + 4(3x²)) = 54x⁵ + 72x³.

17. A First, rewrite the integral: = ∫ tan x dx. You can either derive the integral from using u-substitution, or you should have memorized that ∫tan x dx = −ln|cos x| + C.

18. E Using vertical slices, the area between the curves follows the general formula: . First, it is important to determine where these two curves intersect and which curve is “on top.” The two curves intersect at x = –1, x = 0, and x = 1. Over the interval [–1, 0], the curve y = x³ is “on top.” Therefore, the integral is . Over the interval [0, 1], the curve y = x is “on top.” Therefore, the integral is . Thus the area between the curves is the sum of these two areas: .

19. B Since you are not told which method to use to find the volume you must decide, a big hint is the answer choices. However, if you didn’t have this hint, then you can use the rule of thumb that it is TYPICALLY (but not always) better to use cylindrical shells, if the region is bound by more than two curves (including an axis) or if one or more curves are given as y = and the others are given as x =. Both conditions are satisfied in this problem, so cylindrical shells is probably best. The general formula for cylindrical shells is 2π x[f(x) − g(x)]dx. First, the points of intersection between all these curves must be found, where the region is bound, to establish the limits of integration. The bounds are x = 5 and x = 10. Next, determine which curve is “on top” or “more positive.” In this case, the curves in question are y = (x – 5)³ and y = 0. Since y = (x – 5)³ is always more positive than y = 0, y = (x – 5)³ = f(x) and y = 0 = g(x). Finally, the general formula is for a region that is rotated about the y-axis, or x = 0. Since our curve is shifted to be rotated around x = 2, the radius of the cylinder, x, is now x – 2 to account for the shift. Thus, the final integral is: 2π (x − 2)(x − 5)³ dx.

20. B Use u-substitution in which u = x³ – 3 and du = 3x² dx. Thus, the integral is:

21. B In order to determine the equation for the normal line, take the derivative with respect to x at (2,0): = 6x − 6 which is 6 at (2,0). Since this is the slope of the tangent line, and the slope of the normal line is the opposite reciprocal, the slope of the normal line is −. Plug this information into the point-slope formula of a line and simplify to slope-intercept form: y − 0 = −(x − 2) or y = −x + .

22. E Either use L’Hôpital’s rule or recall that and . In this case, can be rewritten as .

23. D Following the Second Fundamental Theorem of Calculus,

(2 − t²)dt = [2 − (x³)²](3x²) = (2 − x⁶)(3x²) = 6x² − 3x⁸.

24. B Use u-substitution and recognize that the solution will be an inverse sine function. . Thus, the final solution is: + C.

25. D f′(x) = 3x² + 2 and f′(1) = 5.

26. B The first derivative is = 3x² + 4x. The second derivative is = 6x + 4.

27. E For an absolute maximum, find the first derivative and set it equal to zero to determine the critical points. = 20x⁴ − 20x = 0. The critical points are at x = 0 and x = 1. Find the second derivative to determine which of these points are maxima and which are minima. Maxima are located where the second derivative is negative; minima where the second derivative is positive. = 80x³ − 20. From here, x = 0 corresponds with maximum points. Finally, plug in this value into the original equation to determine which has a higher y-value. At x = 0, y = –8. However, you must remember when determining absolute maxima and minima on a closed interval to always check the endpoints. So, plug in x = –2 and x = 2 into the original equation and determine if they have y-values that are more positive than at x = 0. At x = –2, y = –176 and at x = 2, y = 80. Therefore, x = 2 is the absolute maximum.

28. A Using implicit differentiation, find the first derivative of the equation and solve for : Next, determine the second derivative, but don’t simplify the equation: . Finally evaluate the first and second derivative at (–1,–2). The first derivative at (–1,–2) is 2. The second derivative at (–1,–2) is .

29. D Using the Power and Addition Rules, take the derivative of f(x) and you get D. Remember that π and e are constants.

30. E Rolle’s Theorem states that if y = f(x) is continuous on the interval [a, b], and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there is at least one number c between a and b such that f′(c) = 0. f (x) = 0 at both x = 0 and x = 2. Then, solve f′(c) = 8x³ − 16 = 0. c = 2.

31. D An absolute maximum or minimum occurs when the derivative of a function is zero or where the derivative fails to exist or at an endpoint. First, find the derivative of y, set it equal to zero and solve for x. = 5x² − 2x − 7 = 0, then x = –1 and x = . Determine the y-values corresponding to each of these x-values and at the endpoints, x = –2 and x = 2. The resulting points are , , , and . The maximum is occurs at .

32. C The formula for the area under a curve using right-endpoint rectangles is: A = (y₁ + y₂ + y₃ +…y_n), where a and b are the x-values that bound the area and n is the number of rectangles. Since we are interested in the right-endpoints, the x-coordinates are x₁ = , x₂ = , x₃ = , and x₄ = 2. The y-coordinates are found by plugging these values into the equation for y, so y₁ = 3.5625, y₂ = 4.25, y₃ = 5.0625, and y₄ = 6. Then, A = (3.5625 + 4.25 + 5.0625) = 4.71875.

33. D The formula for the area under a curve using inscribed trapezoids is: A = (y₀ + 2y₁ + 2y₂ +…+ 2y_n−1 + y_n), where a and b are the x-values that bound the area and n is the number of rectangles. The x-coordinates are x₀ = 1, x₁ = , x₂ = , x₃ = , and x₄ = 2. The y-coordinates are found by plugging these values into the equation for y, so y₀ = 3, y₁ = 3.5625, y₂ = 4.25, y₃ = 5.0625, and y₄ = 6. Then, A = (3 + 2(3.5625) + 2(4.25) + 2(5.0625) + 6) = 4.34375.

34. E Use the Fundamental Theorem of Calculus: f(x) dx = F(b) − F(a) and u-substitution. For this problem, u = x² and du = 2x dx. Then:

35. B This is an accumulation problem, so use the Second Fundamental Theorem of Calculus, f(t) dt = f(x). Thus, F(x) = = 72 − 0.75 = 71.25.

36. D First, rewrite 2x + y = k in slope-intercept form: y = –2x + k; thus, the slope of the tangent, or the first derivative of y, is –2. Next, take the derivative of y = 2x² – 8x + 14 and set it equal to –2: = 4x − 8 = −2. Use this equation to solve for the x-value that corresponds to the first derivative of –2; the x-coordinate is . Use this x-coordinate to solve for the corresponding y-coordinate; plug x = into the equation for the curve y. The y-coordinate equals . Finally, go back to the equation for the tangent line and plug in the x and y values found. Solve for k: k = .

37. E Since, f′(x) < 0, the curve is decreasing, which is true in all answer choices. In addition, because f″(x) > 0, the curve is concave up. The best way to determine which curve is concave up is to make a quick sketch of the graphs from the data points. When this is done, E is concave up.

38. B The particle slows down when the velocity and acceleration have different signs. First, take the first derivative of x(t) and set it equal to zero to solve for the times when the velocity is changing sign: x′(t) = 6t² − 21t + 9 = 0 when t = and t = 3. Next, take the second derivative of x(t), and determine when that is changing sign: x″(t) = 12t − 21 = 0 when x = . Since there are three different values for which either the velocity or acceleration equals zero, there are four intervals to check the signs of both the velocity and acceleration:

Since the velocity and acceleration have different signs over the intervals and 0 < t < and < t < 3, the correct answer is B.

39. A First, determine where f(x) and the x-axis intercept, i.e., solve f(x) = 0 for x. Thus, x = 0, x = –2, and x = 2. In order to determine the area under the curve, we must set up and solve two integrals: .

40. B The line y = c is a horizontal asymptote of the graph of y = f(x) if the limit of the function as x approaches positive and negative infinity equals c. Similarly, the line x = k is a vertical asymptote of the graph of y = f(x) if the limit of the function as x approaches k from the left and right is positive or negative infinity. First, check for a horizontal asymptote, = 3 and = 3, so there is a horizontal asymptote at y = 3. Next, check for a vertical asymptote; always check the point where the denominator is undefined, in this case, x = –7: and . Thus, there is a vertical asymptote at x = –7.

41. D Use u-substitution in which u = x² – 7 and du = 2x dx. Thus, the integral is:

= ln |x² − 7| + C

42. B The formula for the area between two curves is (f(x) − g(x)) dx, where a and b are the x-coordinates that bind the region and f(x) is the more positive curve. Be careful to check if the curves cross the x-axis because multiple integrals will be required if this happens. In this case, the curves intersect on the x-axis at x = 2. Therefore, the area between the curves will be:

43. B Using the MVTI, = 12.

44. D Plug in the given point, (0,3), into the equation for the curve, y = ax² + bx + c, thus c = 3. Next, rewrite the equation for the normal line in slope-intercept form, y = x + 3. Since this line is normal to the curve, the slope of the tangent line is the opposite reciprocal to the slope of the normal line. The slope of the tangent line is –5 to evaluate it at (0,3). Take the derivative of y and set it equal to –5: = 2ax + b and = 2(a)(0) + b = −5. Therefore, b = –5. Finally, solve for a by plugging the second point (1,5) into the equation for the curve. Also, plug in b and c, so: 5 = a + b + c or a = 7, when b = –5 and c = 3. Finally, the equation of the curve is y = 7x² – 5x + 3.

45. B ((x³ + 2x² − 3)(x⁻² + 7x)) = (x³ + 2x² − 3) • (x⁻² + 7x) = 0 • 8 = 0

1. A cylindrical drum is filling with water at a rate of 25π in³/sec.

(a) If the radius of the cylinder is the height, write an expression for the volume of water in terms of the height at any instance.

(a) V = πr²h and r = . Thus, the volume can be found by solving V = .

(b) At what rate is the height changing when the height is 10 in?

(b) The rate the height is changing can be found by taking the first derivative with respect to time. . Plug in the values given and solve for in/sec.

(c) What is the height of the water when it is increasing at a rate of 12 in/sec?

(c) Use the derivative from part (b) and plug in the values to solve for h, so h = in.

2. The function f is defined by f(x) = (9 − x²) for −3 ≤ x ≤ 3.

(a) Find f′(x).

(a) f′(x) = (9 − x²)(2x) = 3x(9 − x²)

(b) Write an equation for the line tangent to the graph of f at x = –2.

(b) Use the equation for f′(x) from part (a) to find the slope of the tangent at x = –2: f′(−2) = −6. Determine the y-coordinate that corresponds with x = –2 by plugging it into f(x): f(−2) = 5. Finally, the equation for the tangent line is: y − 5 = −6(x + 2).

(c) Let g be the function defined by g(x) = . Is g continuous at x = –2? Use the definition of continuity to explain your answer.

(c) The definition of continuity states a function is continuous if three conditions are met: f(c) exists, f(x) exists, f(x) = f(c). The first condition is met, f(–2) exists, from part (b). The second condition is violated; the left and right hand limits as x approaches –2 are not equal, so the limit does not exist and the function is not continuous: g(x) = 5 and g(x) = 5.

(d) Find the value of

(d) Solve using u-substitution, in which u = 9 – x² and du = 2x dx. Thus, dx = 145.8.

3. A particle moves with velocity v(t) = 9t² + 18t – 7 for t ≥ 0 from an initial position of s(0) = 3.

(a) Write an equation for the position of the particle.

(a) The position function of the particle can be determined by integrating the velocity with respect to time, thus s(t) = ∫v(t) dt. For this problem, s(t) = ∫(9t² + 18t − 7) dt = 3t³ + 9t² − 7t + C. Since we are given the initial position, s(0) = 3, plug that in to solve for C. Thus, C = 3 and the equation for the position of the particle is s(t) = 3t³ + 9t² – 7t + 3.

(b) When is the particle changing direction?

(b) The particle changes direction when the velocity is zero, but the acceleration is not. In order to determine when those times are, set the velocity equal to zero and solve for t. v(t) = 9t² + 18t – 7 = 0 when t = and t = . Since, the time range in question is t ≥ 0, we can ignore t = − . Then, take the derivative of the velocity function to find the acceleration function, as (v(t)) = a(t). For the given v(t), a(t) = 18t + 18. Check that the acceleration at time t = is not zero by plugging into the acceleration function: a(t) = 30. Therefore, the particle is changing direction at t = because v(t) = 0 and a(t) ≠ 0.

(c) What is the total distance covered from t = 2 to t = 5?

(c) The distance covered is found by using the position function found in part (a). Determine the position at t = 2 and subtract it from the position at t = 5. From part (b), we know that the object does not change direction over this time interval, so we do not need to find the time piecewise. Thus, s(5) – s(2) = 568 – 49 = 519.

4. Let f be the function given by f(x) = −2x⁴ + 6x² + 2.

(a) Find the equation for the line normal to the graph at (1, 6).

(a) The line normal to the graph will have a slope that is the opposite reciprocal of the tangent line at that point. Therefore, begin by finding the slope of the tangent line, i.e., the first derivative. f′(x) = −8x³ + 12x and f′(1) = 4. The slope of the normal line is –. So, the equation of the normal line is: y − 6 = −(x − 1).

(b) Find the x and y coordinates of the relative maximum and minimum points.

(b) The relative maximum and minimum will occur at the points when the first derivative is zero or undefined. In this case, set the first derivative to zero and solve for x: f′(x) = −8x³ + 12x = 0, and x = 0, x = , and x = − . To determine which of these is a relative maximum and which is a relative minimum, find the second derivative at each of these critical points. f″(x) = −24x² + 12, and f″(0) = 12, = − 24, and = − 24. By the second derivative test, x = 0 corresponds with a relative minimum because f″(0) > 0 and x = , and x = − correspond with relative maximums because and < 0. To determine the y coordinates of these points, plug them back into f(x): f(0) = 0, and . So, a relative minimum occurs at (0,2) and relative maximums occur at and .

(c) Find the x and y coordinates of the points of inflection.

(c) Points of inflection occur when the second derivative equals zero. Take the second derivative from part (b) and solve for the x values when f″(x) = 24x² + 12 = 0. So, x = and x = − . To determine the y-coordinates for these points of inflection, determine f(x) at each of these points: and . So the points of inflection occur at and .

5. Consider the curve given by x³y² – 5x + y = 3.

(a) Find .

(a) Use implicit differentiation: 2x³y + 3x²y² − 5 + = 0. Simplify and isolate : .

(b) Find .

(b) Use the derivative from part (a) and differentiate again using implicit differentiation: . Replace the value for from part (a) for the final solution:

No need to simplify.

(c) Find the equation of the normal lines at each of the two points on the curve whose x-coordinate is –1.

(c) From the original equation, at x = –1, y = –2 or y = 1. Plug those values into from part (a) to get the slope of the tangents at those points: − and –2, respectively. The slopes of the normal lines are the opposite reciprocals of those slopes, so and , respectively. The equations for the normal are then found by using the point-slope formula. The equations are y + 2 = (x + 1) and y − 1 = (x + 1).

6. Let f be the function given by f(x) = 3x³ – 6x² + 4x.

(a) Find an equation for the normal line at x = 2.

(a) First determine what f(2) equals: f(2) = 8. Next, take the first derivative of f(x): f′(x) = 9x² − 12x + 4. At x = 2, f′(x)(x) = 16. The slope of the normal line would then be −. The equation of the normal line is y − 8 = − (x − 2).

(b) Where are the relative maxima and minima of the curve, if any exist? Verify your answer.

(b) Relative maxima and minima exist where the first derivative is zero or does not exist. Set the first derivative equal to zero and solve or x, f′(x) = 0 at x = . To determine whether this point is a maximum or minimum, check the second derivative. If the second derivative is negative, the point is a maximum. If the second derivative is positive, the point is a minimum. However, if the second derivative is zero, the point is a point of inflection. At x = , f″(x) = 0, so the point is a point of inflection. This curve has no maxima or minima.

(c) Where are the points of inflection? Verify your answer. If there are none, explain why.

(c) It was discovered in part (b) that the point of inflection is at x = . Plug that into f(x) to find the corresponding y-value, which is . Thus, the point of inflection is located at . Justification is the same as the justification in part (b).

1. D .

2. A .

3. B

4. C .

5. E Jump discontinuities exist when the left and right hand limits of the function are not equal at a certain value. A point discontinuity exists when the limit of the function at a certain value does not equal the function at that value. An essential discontinuity is a vertical asymptote. A removable discontinuity occurs where a canceled factor in a rational expression existed. For this function, f(x) = , since the factor (x – 9) can be factored out, there will be a removable discontinuity at x = 9. Also, since the function does not exist at x = 3, there is a vertical asymptote there, so there is both a removable and an essential discontinuity on this function’s graph.

6. E First, the function can simplify to f(x) = . Use the quotient rule and be prepared to simplify to make the multiple derivatives easier to evaluate. The first derivative is f′(x) = . The second derivative is f″(x) = . The third derivative is f′″(x) = . Finally, the fourth derivative is f⁽⁴⁾(x) = .

7. D Use u-substitution and remember your derivatives of trig functions.

= (5)(sec 5x tan 5x) = sec 5x tan 5x.

8. E First, determine where the two curves intersect. Set them equal to each other and solve for x. Thus, x = 0 and x = –1 at the points of intersection that bind the area. (Determine which curve is “on top” in order to solve properly.) Next, integrate using cylindrical shells from x = 0 to x = –1: 2π x(6x² − x − (x² − 6x)) dx = . The washer method would have worked, but you would have had to convert the equations into “x =” form. Recall, it is generally best to use the cylindrical shells method when the axis of rotation is in “x =” form and the equations in “y =” form. When everything is in “y =” form the washer method is generally better.

9. A Since u-substitution will not work, you must integrate by parts. Recall, the formula is ∫u dv = uv − ∫v du. For this problem, u = ln 2x, du = , dv = x³ dx, and v = . (If you are unsure of the derivative of ln 2x, remember the product rule of logarithms and then differentiate.) Now, the integral is ∫x⁵ ln 2x dx = ln 2x − ∫x³ dx = (4ln(2x) − 1) + C.

10. E Use partial fractions to solve this problem. When this is done, the fraction in the integrand becomes . If those two fractions are used as integrands, via u-substitution, the solution is −ln|x + 3| + ln|2x − 1| + C.

11. C By the second derivative test, when the second derivative equals zero, there is a point of inflection at that point. A point of inflection signifies a change in the concavity of the graph of the original function. In addition, if the second derivative is negative over an interval, the graph of the original function is concave down over that interval. If the second derivative is positive, the graph is concave up.

12. A Rather than deal with the chain, product, and quotient rules, use logarithmic differentiation:

13. B Using the Power and Addition Rules, take the derivative of f(x) and you get B. Be careful, C is the integral.

14. C Using the Quotient Rule, , take the derivative of f(x) and you get C.

15. A Using the Chain Rule, , and d (sin x) = cos x and d (cos x) = –sin x, the answer is A.

16. E Use implicit differentiation to find :

3x² + 6xy + 3x² + 3y² + 6xy + 3y² = 0

Isolate terms containing by moving all terms that don’t contain to the right side of the equals sign:

3x² + 6xy + 3y² = −3x² − 6xy − 3y²

Factor out :

(3x² + 6xy + 3y²) = −3x² − 6xy − 3y²

Isolate and simplify:

= −1

17. A The volume of a cone is found from the formula V = πr²h. Because we are told the radius of the cup is 6 in and the height is 18 in, then at any level, the height of the water will be three times the radius, thus, h = 3r. Using this relationship, r can be replaced with in the formula for volume, so V = . Then, to determine the rate that the water level is rising, we must differentiate both sides with respect to t: . Insert the rate of the cup filling and the water level at the instant of interest, 6 in, and solve for . Thus, in/s.

18. B When y = log_a u, . For this problem, u = (x² + 2)³ and = 6x(x² + 2)² (from the Chain Rule). Therefore, f′(x) = .

19. B Since you are relating a sine function and a cosine function, look for a trig identity that easily relates those. In this case, cos²t + sin²t = 1. When you replace cos²t and sin²t with x and y, respectively, the equation becomes x + y = 1. Solve for y to get y = –x + 1.

20. B Recall that the velocity function is the first derivative of a position function with respect to time and the acceleration function is the second derivative of a position function with respect to time. Further, when the velocity and acceleration of a particle have the same sign, the speed is increasing. Thus, to solve this problem, first the position function must be differentiated with respect to time: v(t) = 3t² – 36t – 84. This function is set equal to zero and the critical values, or the times when velocity is equal to 0, are found: t = –2 and t = 14. As the problem states, we are only interested in t ≥ 0, so we can ignore t = –2. Next, find the sign of the velocity over the time ranges of interest: 0 ≤ t < 14 results in v(t) < 0, and t > 14 results in v(t) > 0. Now to determine when the speed is increasing, differentiate the velocity with respect to time to get the acceleration function: a(t) = 6t – 36. Determine when the acceleration is 0 (t = 6) and then find the sign of the acceleration around that time: 0 ≤ t < 6 has a(t) < 0 and t > 6 has a(t) > 0. The times when both the velocity and the acceleration have the same sign is when 0 ≤ t < 6 and when t > 14.

21. A When y = e^u, . For this problem, u = sin²x and = 2 sin x cos x (from the Chain Rule). Therefore, f′(x) = 2e^sin2x sin x cos x.

22. D The formula for the area under a curve using a left-hand Reimann sum is: A = (y₀ + y₁ + y₁ + y₃ +…+ y_n), where a and b are the x-values that bound the area and n is the number of rectangles. Since we do not have evenly spaced subintervals, we must find the width of each subinterval individually and multiply it by the left-endpoint y-value, so A = 1(2) + 3(4) + 7(1) + 5(2) + 3(3) + 6(4) = 64.

23. E Recall, = ln |u| + C, so using u-substitution, u = 3x³ + x and du = (9x² + 1). Then, = 2 ln |u| + C = 2 ln |3x³ + x| + C.

24. A This is a complicated integral and integration by parts is the best way to go for solving it. Recall, the formula for integration by parts is ∫udv = uv − ∫vdu. For this problem, let u = e^x and dv = sin x dx. From these, du = e^x dx and v = –cos x. When you input these using the integration by parts formula you get: ∫e^x sin x dx = −e^x cos x + ∫ cos xe^x dx. Since we cannot readily integrate ∫cos xe^x dx, we must use integration by parts again. In this case, u = e^x and dv = cos x dx. From these, du = e^x dx and v = sin x. Insert these in to the equation in place of ∫ cos xe^x dx and you get: ∫e^x sin x dx = −e^x cos x + e^x sin x − ∫e^x sin x dx. We are back to where we started. However, we are in a good position now, because we can add ∫e^x sin x dx to both sides and we end up with 2∫e^x sin x dx = −e^x cos x + e^x sin x. In order to solve for ∫e^x sin x dx, just divide both sides by 2: ∫e^x sin x dx = + C.

25. A All the options are improper integrals and they will converge if their limits as they approach infinity exist. So, check them one at a time and POE. I. , so I. converges. II. = undefined, so II. diverges. III. = infinity, so III. diverges.

26. E Area under polar curve is found by using: A = r² dθ. This curve repeats after 2π, thus the region is bound by 0 and 2π. The area is then found from: A = (5 + 2 cosθ)²dθ. Solve this and the area is found to be 27π.

27. A Solve the differential equation by separating the variables and integrating. The equation would then become: ∫y² dy = ∫cos x dx. From here, y³ = 3 sin x + C. Next, use the initial condition (0, –1) to get the exact equation: y³ = 3 sin x – 1.

28. D A series converges to L when a_n = L. There are many tests that can be used to test whether a series will converge. The test depends on the type of series. So, check each series in this problem one at a time and POE. I. is a geometric series. Geometric series converge if |r| < 1 and the general form of the series is arⁿ⁻¹. The r in I. is , so the series converges. II. can be tested using the ratio test which states that a series in the form a_n converges if < 1. Then II. can be tested as follows, . Since is less than 1, the series converges. For III., the integral test can be used. The integral test states a_n and f(x) dx either both converge or both diverge, for a_n = f(n). Therefore, evaluate the improper integral . This integral diverges (it equals infinity), so the series diverges.

29. B The balloon losing air at a rate of −24π in³/sec means the volume is shrinking at that rate, but it does not directly relate to surface area. Therefore, we must find an intermediate rate to relate the two formulas, V = πr³ and A = 4πr². The common value is r, so the common rate will be . To solve for , differentiate the formula for V with respect to t: . Insert the values for and r and solve for . Thus, in. Next, differentiate the formula for A with respect to t: . Insert the values for and r to solve for . Then, = −24π in²/sec.

30. A When you insert 0 for x, the limit is , which is indeterminate. Use L’Hôpital’s Rule to evaluate the limit: . Since this limit is also indeterminate, use L’Hôpital’s Rule again: . This limit exists and equals .

31. C The horizontal and vertical components of the velocity of a curve are found parametrically as the derivatives with respect to time of the x and y components of the motion, respectively. For this curve, = 5t and = 6t² − 2t + 1. When those two expressions are set equal to each other and solved for t, t = 1 and t = . Since t = 1 is not an answer choice, C is the correct answer.

32. D Use implicit differentiation to find :

2x + 6xy + 3x² + 3y² + 6xy + 3y² = 0

Do not rearrange the terms to isolate . Instead, plug in (3,2) immediately, solve for the derivative, and simplify:

2(3) + 6(3)(2) + 3(3)² + 3(2)² + 6(3)(2) + 3(2)² = 0

6 + 36 + 27 + 12 + 36 + 12 = 0

54 + 75 = 0

33. E

From the diagram, it is clear D₁ = and D₂ = 1500 − x. Since distance = rate • time and r₁ = 3 and r₂ = 5, the times for each leg of the trip are t₁ = and t² = . Therefore, the total time to ride home is T = + 300. To minimize the time, the derivative of T is taken and set equal to zero. and x = ±. The negative value can be ignored. Because the derivative is ugly, it will take too much time to check the second derivative. (Don’t make skipping this a habit, though!) Therefore, the time is minimized when x = m. The kid should cross at 1462.5 m from his house.

34. B The profit is maximized at the values that make the derivative of the profit equation equal to zero or at the end points of the range. The derivative of the profit equation is = 6x² − 210x − 1500 = 0. This equation is true when x = 10 or 25. Then, those values and the endpoints, x = 0 and x = 50 are used to solve for P. The resulting points are (0, –1200), (10, 5300), (25, 1925), and (50, 61300). Since the profit equation is maximized at x = 50, and the profit is in millions of dollars, the final result is $61.3 billion.

35. B Use the mean value theorem for integrals, f(c) = f(x)dx. Thus, for this problem, 5 = x⁴ dx. Using the Fundamental Theorem of Calculus and solving the equation for k, k = ±.

36. A Recall, ∫a^u du = + C. In this problem, u = 3x² + 2 and du = 6 x dx.

Thus, ∫7^u du = + C

37. B The area of a rectangle is A = lw. In this case, the width is 2x and the length cos x + 1, so A = 2x (cos x + 1). To minimize the area between the rectangle and the graph, the area of the of the rectangle will be maximized. To maximize the area of the rectangle, you must take the derivative of the area and set it equal to zero: = 2 cos x − 2x sin x + 2 = 0. When you solve this with your calculator, there will be four critical points: −π, −1.30654, 1.30654, π. Plug the points into the formula to determine the area at each point. At –π and π, the area is 0. At −1.30654 and 1.30654, the area is 3.29559. (You can confirm this is the maximum by taking the second derivative and checking that it is negative at 1.30654.) Next, determine the area under the curve by using a definite integral. cos x + 1dx = 2π. To determine the area between the rectangle and the curve, subtract the area of the rectangle from the area under the curve: A_between = 2π − 3.29559 = 2.98759.

38. D The current state of the integral makes it appear very difficult to solve. However, notice that there is an x² term and no square root. So, think inverse tangent! First, complete the square: x² + 6x + 10 = (x + 3)² + 1. Thus, you can rewrite the integral as . Now, u = x + 3 and du = dx. Use the pattern = tan⁻¹ u + C. Therefore, = tan⁻¹(x + 3) + C.

39. E In order to find the distance the particle travels, set the first derivative equal to zero to determine, when, if at all, it changes direction over the time interval: x′(t) = 4t − 3 = 0, hence t = . Since the particle changes direction at t = , we must find the position of the particle at t = 0, t = , and t = 4. x(0) = 1, , and x(4) = 21. The total distance the particle travels is found by .

40. E Given a table of values, a left-hand Riemann sum can be calculated by multiplying the size of the intervals (i.e., the difference in x-values), by the left endpoint y-values and summing them all up.

41. C Recall L = . In this case, = x(x² − 2) and L = .

42. E From the Second Fundamental Theorem of Calculus, (t − t⁴) dt = x − x⁴.

43. D This is a related rates problem. Since the volume of a cylinder is given by V = πr²h, begin by taking the derivative of both sides of the equation with respect to t. Because the radius is constant, treat it as a constant when taking the derivative: = πr² . Now, plug in the given values for and r: 96π = π4². Solve for = 6.

44. B Draw a picture of the situation:

The question is asking about θ in this right triangle and we are given the measurements of the two legs, so set up your equation using tan: tan θ = . As this is a related rates problem, take the first derivative with respect to t: . Notice the rate of the child rising, , was given in ft/min while the question asks for the solution in rad/sec. Convert into ft/sec: ft/sec. Also, note that a height, h, was given in the problem. Plug that into the original equation for tan to solve for θ or directly solve for sec² θ: tan θ = . Recall, 1 + tan²θ = sec²θ, so plug in to solve for sec²θ: sec²θ = . Now, plug in all these values into the derivative of tan and solve for or rad/sec.

45. C Use the quotient rule and recall the derivative of the natural log and e.

.

1. Let y be the function satisfying f′(x) = −x(1 + f(x)) and f(0) = 5.

(a) Use Euler’s method to approximate f(1) with a step size of 0.25.

(a) There are two equations you need to know for Euler’s method: 1. x_n = x_n−1 + h and 2. y_n = y_n−1 + hy′_n−1. In this case, there will be four steps (n = 4) and h = 0.25. From the given information, y′ = −x(1 + y), so y′₀ = 0. It is also given that x₀ = 0 and y₀ = 5. Then, x₁ = 0.25, y₁ = 5, and y′₁ = 1.5. Next, x₂ = 0.5, y₂ = 4.625, and y′₂ = –2.8125. Once more, x₃ = 0.75, y₃ = 3.92188, and y′₃ = –3.69141. Finally, x₄ = 1 and y₄ = 2.999.

(b) Find an exact solution for f(x) when x = 1.

(b) Solve the differential equation: = −x(1 + y). Then, y = Ce − 1. With the initial condition that f(0) = 5, f(x) = 6e − 1. Therefore, f(1) = 6e⁻ − 1 ≈ 2.639.

(c) Evaluate −x(1 + f(x))dx.

(c) First, remember the Fundamental Theorem of Calculus: f′(x)dx = f(b) − f(a). From part (b), f(x) = 6e − 1. Notice that −x(1 + f(x))dx is an improper integral. So, evaluate f(a) and f(0) and find the difference. f(a) = −1 and f(0) = 5. Therefore, −x(1 + f(x))dx = −6.

2. Let R be the region enclosed by the graphs of y = x² – x – 6 and y = x³ – 2x² – 5x + 6 and the lines x = –2 and x = 2.

(a) Find the area of R.

(a) Area = [f(x) − g(x)]dx. We are told the region is bound by x = –2 and x = 2. Further, y = x³ – 2x² – 5x + 6 is always more positive or “on top” over the entire region, so f(x) = x³ − 2x² − 5x + 6 and g(x) = x² – x – 6. Now, the integral can be set up: A = (x³ − 2x² − 5x + 6) − (x² − x − 6)dx. When solved, the area is found to be 32 units squared.

(b) The horizontal line y = 0 splits R into two parts. Find the area of the part of R above the horizontal line.

(b) The line y = 0 divides R so that the area above R is the region enclosed by y = 0 and y = x³ – 2x² – 5x + 6. Now, the limits of integration are when these two curves intersect. These curves intersect at x = –2, x = 1, and x = 3. However, R is bound between x = –2 and x = 2, so the limits of integration are x = –2 and x = 1. The same formula for area can be used from part (a), but the specific values input would be the ones described here. Thus, A = (x³ − 2x² − 5x + 6 − 0)dx. Thus, the area of this portion of R = .

(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an equilateral triangle. Find the volume of this solid.

(c) The volume of the solid is found by integrating the area of the cross-section, so V = A(x)dx. In this case, since the cross-section is an equilateral triangle, A = (side)³, where the side length is the length between the curves, i.e., f(x) – g(x). From part (a), we know and g(x) = x² – x – 6 and g(x) = x² − x − 6 and the limits of integration are x = –2 and x = 2. The equation for the volume is then V = [(x³ − 2x² − 5x + 6) − (x² − x − 6)]dx = = 141.467 units cubed.

(d) What is the volume of the solid generated by the region R, when it is revolved about the line x = –3.

(d) Since you are not told which method to use to find the volume you must decide. You can use the rule of thumb that it is typically (but not always) better to use cylindrical shells, if the region is bound by more than two curves (including an axis) or if one or more curves are given as y = and the others are given as x =. Both conditions are satisfied in this problem, so cylindrical shells is probably best. The general formula for cylindrical shells is 2π x[f(x) − g(x)]dx. We know from part (a) the limits of integration are x = –2 and x = 2 and we have established which curve is f(x) and which is g(x) in part (a). The only thing left to establish is the radius of the cylinder from the axis of rotation. Since there is a shift in the axis to x = –3, the radius is x + 3. Thus, the final integral is: V = 2π (x + 3)[(x³ − 2x² − 5x + 6) − (x² − x − 6)]dx = = 542.688 units cubed.

3. The derivative of a function f is f′(x) = (2x + 6)e^−x and f(2) = 15.

(a) The function has a critical point at x = –3. Is there a relative maximum, minimum, or neither at this point on f? Justify your response.

(a) You have two options here: 1) Using the first derivative test: Since x = –3 is a critical point, a local minimum will exist when f′(x) < 0 for x < –3 and f′(x) > 0 for x > –3. In this case, those two criteria are satisfied, so x = –3 is a relative minimum. 2) You can use the second derivative test: Take the derivative of f′(x). f″(x) = (−2x − 4)e^−x. Plug x = –3 into f″(x). From the second derivative test, if f″(c) < 0, then c is a relative maximum, but if f″(c) > 0, then c is a relative minimum. In this case, f″(−3) = 2e⁻³ which is greater than 0, so x = –3 is a relative minimum of f.

(b) On what interval, if any, is the graph of f both increasing and concave down? Explain your reasoning.

(b) First, use the second derivative from part (a). Find the points of inflection by setting f″(x) = 0. There is a point of inflection at x = –2. Now, test the behavior of f″(x) around x = –2. When x < –2, f″(x) > 0, which means the curve is “concave up.” When x > –2, (x) < 0, which means the curve is “concave down.” To determine when the curve is decreasing, the first derivative must be tested. We already know from the question stem to part (a) that there is a critical point at x = –3. So, test the behavior of f′(x) around x = –3. For x < –3, (x) < 0, and, thus, falling or decreasing. For x > –3, f′(x) > 0 and, thus, rising or increasing. When the behaviors of the first and second derivatives of f are combined, f(x) is increasing and concave down over the interval x > –2.

(c) Find the value of a f(5).

(c) In order to find f(5), you can integrate to find f and then use the initial condition f(2) = 15 to find the exact solution. Once the exact solution (the exact equation) is found, then f can be evaluated at x = 5. A more direct path would be to integrate (x) over the range from x = 2 to x = 5. The value found can then be added to f(2). So the solution would be f(5) = f(2) + (2x + 6)e^−x dx = 15 − 18e⁻⁵ + 12e⁻².

4. Consider the equation x³ – 2x²y + 3xy² – 4y³ = 10.

(a) Write an equation for the slope of the curve at any point.

(a) Find the first derivative of this equation using implicit differentiation:

(b) Find the equation of the normal line to the curve at the point x = 1.

(b) At x = 1, y = –1. Plug these values into the equation for from part (a). Then, the slope of the tangent line is 1. The slope of the normal line is the opposite reciprocal, which is –1. Using point-slope formula, the equation of the normal is y + 1 = x – 1 or y = x – 2.

(c) Find at x = 1.

(c) The second derivative follows directly from the first in part (a):

Rather than trying to simplify the equation, plug in the value for from part (b) and x = 1 and y = –1. When simplified using those values, = .

5. Given that f(x) = sin x:

(a) Find the 6^th degree Maclaurin series.

(a) In order to determine the 6^th degree Maclaurin series, use the general formula: . So, find f(x) and the first 5 derivatives of f(x):

f(x) = sin x, f′(x) = cos x, f″(x) = −sin x, f″′(x) = −cos x, f⁽⁴⁾(x) = sin x, and f⁽⁵⁾(x) = cos x

Then, evaluate each of those equations at x = 0:

f(0) = 0, f′(0) = 1, f″(0) = 0, f″′(0) = −1, f⁽⁴⁾(0) = 0, and f⁽⁵⁾(0) = 1. Finally, insert the values into the formula for a Maclaurin series and simplify: sin x = .

(b) Use the polynomial to estimate sin 0.2.

(b) From part (a), sin x = x − . To approximate sin 0.2, plug 0.2 in for x in the equation and simplify: sin 0.2 ≈ 0.2 − ≈ 0.201336.

(c) Estimate the remainder of the approximation.

(c) The formula for the Lagrange remainder is R_n(x, a) ≤ f⁽ⁿ⁺¹⁾(c). Use the series you found in part (a) to generalize the series sin x = . From here, R_2n(x, 0) ≤ . (in this case, it is R_2n not R_n because every other term is 0.) Recall, |sin x| ≤ 1, so R_2n+1(x,0) ≤ . Therefore, R₇(x, 0) ≤ . (In general, a good approximation of the remainder/error bound of an nth degree Taylor polynomial is the next nonzero term in a decreasing series.) For the approximation in part (b), plug 0.2 for x into . Therefore, the remainder is ≈ 2.540 × 10⁻⁹.

6. Two particles travel in the xy-plane for time t ≥ 0. The position of particle A is given by x = 2t – 3 and y = (2t + 1)² and the position of particle B is given by x = t – 1 and y = t + 23.

(a) Find the velocity vector for each particle at t = 3.

(a) The velocity vector of each particle is found by taking the derivative of each particle’s motion, so the vector is . For particle A, = 2 and = 8t − 4. For particle B, = 1 and = 1. Evaluate each of these derivatives at t = 3, so the final velocity vectors are A = (2,20) and B = (1,1).

(b) Set up, but do not evaluate, an integral expression for the distance traveled by particle A from t = 3 to t = 5.

(b) The formula for the distance travelled parametrically, or the length of the curve, is . Since the velocity vector does not equal zero at any time over the interval from t = 3 to t = 5, the particle does not change direction and we can plug in the formulas for the x and y components of the velocity from part (a): L = , which is simplified to L = .

(c) At what time do the two particles collide? Justify your answer.

(c) The particles will collide when the x and y coordinates of their position functions are equal to each other. Therefore, set the position functions equal and solve for t. Beginning with the x-components: 2t – 3 = t – 1, so t = 2. Plug t = 2 into the y-components and confirm they are equal. For particle A: y = (2 • 2 + 1)² = 25. For particle B: y = 2 + 23 = 25. Both the x and y components are equal at t = 2, so that is when they collide.

1. D To do this question, we need to remember how to attack limits. These are generally good candidates for the first pass. When the limit approaches infinity, simply divide both top and bottom by the highest power of x in the fraction. Foiling out the bottom produces a 2x⁴ as the leading term, so both top and bottom will be divided by an x⁴ term.

Simplify.

As x gets larger, the fractional terms approach 0 so all that is left is which equals 2.

Also, when limits approach infinity, it is good to keep in mind that when the degrees are the same in the numerator and denominator, then you only need to look to the leading coefficients to find the limits.

2. A When calculating limits approaching ∞, you should pay attention to the highest degree of the numerator and denominator. If the highest degree is in the numerator, then the limit will become infinitely larger without bound, thus making it ∞. The long way is to divide each term by the highest term, in this case, x⁵.

Simplify

As x approaches ∞, the fractional terms will get smaller and smaller, going to 0. The only value left is in the numerator and as x would get larger and larger, the limit increases without bound, making it ∞.

3. B When we look to the highest degree, it is in the denominator, which, when limits approach infinity, automatically means the entire expression will approach 0.

Divide by the highest term throughout the expression.

Simplify.

4. B Know the rule: When a limit approaches infinity and the degrees in the numerator and denominator are the same, simply look to the leading coefficients to calculate the limit. In this case, the leading coefficients are the 3 in the numerator and the 6 in the denominator . reduces to , making it your answer.

5. D This question tests your knowledge of discontinuity. If you plug in –3 directly, the denominator becomes 0, and thus is undefined. If we factor and cancel the numerator and denominator, the problem becomes solvable.

Factor and cancel out like terms.

Plug in –3 to evaluate the limit.

6. D This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If we expand out the numerator and reduce, the problem becomes solvable.

Expand the numerator.

= = =

Plug in 0 to evaluate the limit.

0² − 3(0) + 3 = 0² − 3(0) + 3 = 3

7. B This question tests your knowledge of discontinuity. If you plug in 16 directly, the denominator becomes 0, and thus is undefined. If we multiply both the numerator and denominator by 4 + , then the question becomes solvable.

Multiply top and bottom by 4 + .

Plug in 16 to evaluate the limit.

8. C This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If we multiply both the numerator and denominator by 1 + , then the question becomes solvable.

Multiply the top and bottom by 1 + .

Plug in 0 to evaluate the limit.

9. B This question tests your knowledge of discontinuity. If you plug in 1 directly, the denominator becomes 0, and thus is undefined. If you factor and cancel like terms, then the question becomes solvable.

Factor and cancel like terms.

Plug in 1 and evaluate the limit.

10. E This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If you expand out the numerator and cancel like terms, then the question becomes solvable.

Expand out numerator and cancel like terms.

Plug in 0 to evaluate the limit.

x + 14 = 0 + 14 = 14

11. A To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In this case, they are both 10, so the limit is simply the coefficients divided by each other, , or 7.

12. C To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In this case, they are both 2, so the limit is simply the coefficients divided by each other, .

13. D To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In the numerator, the degree is 2, where in the denominator, the degree is 3. When the degree in the denominator is greater than the degree in the numerator, the limit will approach 0.

14. B The first instinct in limit questions is to try to plug in the value x is approaching. If you plug 3 in, the denominator becomes 0, thus undefined. However, using a little algebra, we are able to remove a term from the numerator and denominator.

Now that the numerator and denominator are simplified, you can simply plug 3 in for x to evaluate the limit.

15. E If you plug in 0 directly, the value becomes and can’t be evaluated. However, an h is able to be cancelled out of the denominator by factoring.

Now that the numerator and denominator are simplified, you can simply plug 0 in for h to evaluate the limit.

16. C If you plug in –2, the limit becomes , which is undefined. By factoring the denominator, the discontinuity can be removed which will allow calculation.

Now the limit is able to be evaluated by plugging in –2 for x.

17. A If you plug in 5 immediately, the limit becomes , which is undefined. By factoring the numerator, the discontinuity can be removed which will allow calculation.

Now the limit is able to be evaluated by plugging in 5 for x.

−(5)(5) = −25

18. B If you plug in 7 directly to the limit, then it becomes which is undefined. First, the numerator must be factored to remove the discontinuity.

Plug in 7 to evaluate the limit.

−(7) + (7) = −(14) = −14

19. E When evaluating limits at infinity, you must pay attention to the degrees in both the numerator and denominator. In this case, the degree in the numerator is 100, while in the denominator, it’s 99. Since the degree is higher in the numerator, the limit will go to infinity as x approaches infinity.

20. D If you plug in 3 directly, the limit becomes and is undefined. First, the numerator and denominator must be factored.

Now, plug 3 in for x to evaluate the limit.

21. B .