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 :
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,
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 +…yn), 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 +…+ 2yn−1 + yn), 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:
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:
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