AP Calculus Chapter 3 Testbank (Mr. Surowski)

AP Calculus Chapter 3 Testbank (Mr. Surowski) Part I. Multiple-Choice Questions (5 points each; please circle the correct answer.). If f(x) = 0x 4 3 + x, then f (8) = (A) (B) 4 3 (C) 83 3 (D) 2 3 (E) 2 2. If g(x) = 3x2 + x 3x 2 x, then g (x) = (A) (B) 6x2 + 6x 2 (C) 6 (3x ) 2 (D) 2x 2 (E) 36x2 2x (x 2 x) 2 (x 2 x) 2 3. If f(x) = + x, then f (x) = (A) 4 x + x (B) 2 x + x (C) 4 + x (D) 4 x + x (E) 2 x + x

4. Find given that x3 y + xy 3 = 0. (A) 3x 2 + 3xy 2 (B) (3x 2 + 3xy 2 ) (C) 3x2 y + y 3 3xy 2 + x 3 (D) 3x2 y + y 3 3xy 2 + x 3 (E) x2 y + y 3 xy 2 + x 3 5. If f(x) = sin 2 x, find f (x). (A) sin 2 x (B) 2 cos 2x (C) cos 2x (D) 4 sin 2x (E) sin 2x 6. If f(x) = 3 πx, find f (x) = 3 πx (A) (B) 3πx (C) 3πx π ln 3 ln 3 π (D) π ( 3 πx ) (E) π ln 3 (3 πx ) 7. Find the slope of the normal line to the graph of y = x + cos xy at the point (0, ). (A) (B) (C) 0 (D) 2 (E) Undefined 8. If f(x) = 3x 2 x and g(x) = f (x), then g (0) could be (A) 59 (B) 59 (C) (D) (E) 0 Note that f(x) = 3x 2 x = x = g(3x 2 x). Differentiate this and get = (6x )g (3x 2 x). One of the solutions of 3x 2 x = 0 is x = 2; for this value get = g (0). 9. If the function f(x) is differentiable and f(x) = a = { ax 3 6x if x bx 2 + 4 if x >, then (A) 0 (B) (C) 4 (D) 24 (E) 26

0. Two particles leave the origin at the same time and move along the y-axis with their respective positions determined by the functions y = cos 2t and y 2 = 4 sin t for 0 < t < 6. For how many values of t do the particles have the same acceleration? (A) 0 (B) (C) 2 (D) 3 (E) 4. Find the value(s) of at y = given that x2 y + y 2 = 5. (A) 3 2 only (B) 2 3 only (C) 3 2 only (D) ±2 3 (E) ± 3 2 2. If f(x) = x 2 3x +, then f (x) = (A) 3x2 2x 3x + (B) 9x2 + 2x 3x + (C) 9x2 + 4x 2 3x + (D) 5x2 + 4x 2 3x + (E) 9x2 4x 2 3x + 3. What is the instantaneous rate of change at t = of the function f, if f(t) = t3 + t 4t +? (A) 2 9 (B) 4 9 (C) 20 9 (D) 4 9 (E) 2 9

4. What is the equation of the line tangent to the graph of y = sin 2 x at x = π 4? (A) y ( 2 = x π ) 4 (B) y ( 2 = x π ) 4 (C) y ( = x π ) 2 4 (D) y = ( x π ) 2 2 4 (E) y 2 = 2 ( x π ) 4 { 3ax 2 + 2bx + if x 5. If the function f(x) = ax 4 4bx 2 3x if x > real values of x, then b = (A) 4 (B) 4 (C) 7 6 is differentiable for all (D) 0 (E) 4 6. The position of a particle moving along the x-axis at time t is given by x(t) = e cos 2t, 0 t π. For which of the following values of t will x (t) = 0? I. t = 0 II. t = π 2 (A) I only (B) II only (C) I and III only (D) I and II only (E) I, II, and III III. t = π

7. If f(x) = (3x) 3x, then f (x) = (A) (3x) 3x (3 ln(3x) + 3) (B) (3x) 3x (3 ln(3x) + 3x) (C) (9x) 3x (ln(3x) + ) (D) (3x) 3x (3x) (E) (3x) 3x (9x) 8. Given that f(x) = 2x 2 + 4, which of the following will calculate the derivative of f(x)? (A) [2(x + x)2 + 4] (2x 2 + 4) x (2x 2 + 4 + x) (2x 2 + 4) (B) lim x 0 x (C) [2(x + x) 2 + 4] (2x 2 + 4) lim x 0 x (D) (2x2 + 4 + x) (2x 2 + 4) x (E) None of the above. 9. Given that g(x) =, which of the following will calculate the derivative x + of g(x)? ( (A) x x + x + ) x + ( (B) lim x 0 x x + x + ) x + ( ) ( (C) lim lim x 0 x x 0 x + x + ) x + ( (D) lim x 0 x + x + ) x + (E) None of the above.

The next two questions pertain to the function f, whose graph is given below: 20. For the function f, I. f ( 3) > 0 II. f (0) < 0 III. f is differentiable on the interval (0, ) (A) I only (B) II only (C) III only (D) I and II (E) I, II, and III y=f(x) 0 y 5 x -6-4 -2 2 4 6-5 -0 2. For the function f I. f (x) > 0 on the interval ( 5, 4) II. f (x) is constant on the interval (4, 6) III. f is not defined at all points of the interval (, 5) (A) I only (B) II only (C) III only (D) I and II (E) II and III

22. Given the graph of the rational function f below, give a sketch of the graph of y = f (x) on the same coordinate axes. (Note: the graph of y = f(x) has a vertical asymptote at x =.) The graph of y = f (x) is in blue. 8 Y 6 4 2-4 -2-2 2 4-4 X 23. The following graph represents a function g defined on the interval [ 4, 4] and differentiable on ( 4, 4). On the same coordinate axes, graph y = g (x) over the interval ( 3, 3). The graph of y = g (x) is in blue. 4 Y 2-4 -2 2 4-2 X -4

24. The following graph is that of y = h (x). On the same coordinate axes, give a sketch of y = h(x), assuming that h(0) =. The graph of y = h(x) is in blue. Y 4 3 2-4 -2-2 4-2 X 25. The following graph is that of y = h (x). On the same coordinate axes, give a sketch of y = h(x), assuming that h(0) = 0. The graph of y = h(x) is in blue. Y X

26. Using the definition of the derivative of a function, find f (x), where f(x) = x x 4. Then find f (). f f(x + h) f(x) (x) = lim h 0 h (x + h) (x + h) 4 (x x 4 ) = lim h 0 h (x + h) (x 4 + 4x 3 h + 6x 2 h 2 + 4xh 3 + h 4 ) (x x 4 ) = lim h 0 h h (4x 3 h + 6x 2 h 2 + 4xh 3 + h 4 ) = lim h 0 h = lim[ (4x 3 + 6x 2 h + 4xh 2 + h 3 )] = 4x 3. h 0 That is to say, f (x) = 4x 3, and so f () = 4 3 = 3. 27. Using the definition of the derivative of a function, find where y = x. Then find. x=4 = lim f(x + h) f(x) h 0 h x + h x = lim h 0 h ( x + h x)( x + h x) = lim h 0 h( x + h + x) = lim h 0 = lim h 0 Therefore, = x=4 2 4 = 4. x + h x h( x + h + x) h h( x + h + x) = lim = h 0 x + h + x 2 x

28. Let f(x) = 4x 3 2x 2 24x + 23. (a) Compute f (x). f (x) = 2x 2 42x 24. (b) Find all values of x satisfying f (x) = 0. f (x) = 0 2x 2 42x 24 = 0 2x 2 7x 4 = 0 (2x + )(x 4) = 0 x = 2, 4. 29. Let f(x) = x + x. (a) Compute f (x). f (x) = x 2 (b) Find all values of x satisfying f (x) = 0. We have 0 = f (x) = x 2 = x2 x 2 x = ±. 30. Let y = x + x 2. (a) Compute. Using the quotient rule, one has d = (x)( + x2 ) x d ( + x2 ) ( + x 2 ) 2 = ( + x2 ) x(2x) ( + x 2 ) 2 = x2 ( + x 2 ) 2. (b) Compute all values of x for which = 0. From the above, it s clear that = 0 x = ±.

3. Let s(x) = sin x x and compute lim x s (x). We have, using the quotient rule, that s (x) = x cos x sin x. Therefore, ( cos x lim x s (x) = lim x x sin x ) = 0 0 = 0. x 2 x 2 32. The graph below depicts the velocity v = s (t) of a particle moving along a straight line, where on this straight line positive direction is to the right. 4 v (velocity) 2-2 2 4 6 8 0 t (time) -4 (a) Would you say that at time t = the particle is moving to the left, moving to the right, or not moving at all? Please explain. Particle is moving to the right, as v() > 0. (b) Would you say that at time t = 3 the particle is moving to the left, moving to the right, or not moving at all? Please explain. Particle is moving to the left, as v(3) < 0. (c) Would you say that at time t = 4 the particle is moving to the left, moving to the right, or not moving at all? Please explain. Particle is not moving, as v(4) = 0. (d) Find (estimate) two values of t at which time the particle is not accelerating. It appears that a(t) = v (t) = 0 where t or 3. At such values of t the particle will not be accelerating.

(e) Find (estimate) a value of t at which time the particle is moving to the left, but with zero acceleration. This would happen at t 3 as a(3) 0 and v(3) < 0. (f) According to this graph, at how many distinct times is the particle at rest? The particle is at rest when it s not moving; this happens for FIVE values of t. (g) For which values of t is the particle not only at rest, but is not accelerating (i.e., has no forces acting on it)? There are NO SUCH values of t. (h) According to this graph, at how many distinct times is the particle not accelerating? We have a(t) = 0 for FOUR values of t. 33. Using logarithmic differentiation compute f (x) where f(x) = x x, x > 0. Starting with ln f(x) = ln x x = x ln x, and differentiating both sides, we get f (x) f(x) = ln x +, and so f (x) = f(x)(ln x + ) = x x (ln x + ). 34. Let P (t) = + e kt, where k is a real number. (a) Show that dp = kp ( P ). dt Using the chain rule, we have dp dt ke kt = ( + e kt ) 2 k = + e e kt kt + e kt = kp ( P ). (b) Show that d2 P = 0 when P = /2. dt2 Using implicit differentiation, together with the result of part (a), we have that d 2 P dp 2 = kp ( P ) kp P = kp ( 2P ). From the above, it s now obvious that d2 P dp 2 = 0 when P = 2.

35. Let f and g be differentiable functions and assume that f() = 2, f () =, g() =, g () = 0. Compute h (), given that h(x) = x 2 f(x)g(x). This uses only the product rule: Substituting x = yields h (x) = 2xf(x)g(x) + x 2 f (x)g(x) + x 2 f(x)g (x). h () = 2f()g() + f ()g() + f()g () = 2 2 ( ) + ( ) + 2 0 = 5. 36. In your text it was given as a exercise that the dollar cost of producing x washing machines is c(x) = 2000 + 00x 0.x 2. Why is this an absolutely rediculous cost model? (What is lim x c(x)? Is this reasonable?) The above model says that the cost of producing x washing machines eventually becomes NEGATIVE. This is clearly preposterous! 37. The volume V is a sphere of radius r is given by the formula V = (4/3)πr 3. Suppose that you know that the radius r is an increasing function of t, and that when r = 3, dr dt = 2. Compute dv dt Using the chain rule, dv dt = 4πr2dr dt when r = 3. r=3 = 4π 3 2 2 = 72π. 38. Compute d ( ) cos x, simplifying as much as possible. + sin x Using the quotient rule, ( ) d cos x = sin x( + sin x) cos2 x = + sin x ( + sin x) 2 ( + sin x) ( + sin x) 2 = + sin x. Exercise 0, page 30.

39. Note that the point (, 2) is on the curve defined by y 3 xy 2 x 2 y 2 = 0. (a) Compute at the point (, 2). Using implicit differentiation, one has 3y 2 y y 2 2xyy 2xy x 2 y = 0, and so and so y = = y 2 + 2xy 3y 2 2xy x 2, y (, 2) = 2 2 + 2 2 3 2 2 2 2 2 = 8 7 (b) Find an equation of the straight line tangent to the above curve at the point (, 2). Such an equation can be written as which becomes y 2 = 8 7 (x ). y 2 = y (, 2)(x ), (c) Find an equation of the straight line normal to the above curve at the point (, 2). Such an equation can be written as y 2 = which becomes y 2 = 7 8 (x ). ( ) (x ), y (, 2) 40. Suppose that y = e x cos x. Show that y + 2y + 2y = 0 This is routine: y = e x cos x e x sin x, and so y = e x cos x + e x sin x + e x sin x e x cos x = 2e x sin x. Therefore, y + 2y + 2y = 2e x sin x + 2( e x cos x e x sin x) + 2(e x cos x) = 0.

4. Find f x (x), given that f(x) = x2 + 9 possible. Using the quotient and chain rules, and simplify your result as much as f (x) = x2 + 9 2x 2 (x 2 + 9) /2 x 2 + 9 = 9 x 2 (x 2 + 9) 3/2. 42. Compute f (), given that f(x) = sin ( π x 2 + 3 ). f (x) = x cos ( π x 2 + 3 ) ; therefore, f () = /2. x2 + 3 43. Let x = t 2, y = t t + (a) Compute in terms of t. = dt / dt = 2 (t + ) 2t = 4t 2 (t + ) 2. (b) Find all values of t for which From part (a) give x and y parametrically in terms of t. fails to exist. fails to exist when t =. (c) Find all values of t for which = 0. From part (a) = 0 when t = 0. (d) Compute lim x, lim y, lim t t t. lim = + ; lim y = ; lim t t t = 0. (e) Suppose that the graph of y = f(x), where f is a differentiable function, has a horizontal asymptote (say with lim y = c, for some real number c. x Would you expect that lim x = 0. This is a bit subtle, but we cannot infer that lim = 0 (even though x we might expect this to happen!). A counterexample is y = sin(x2 ). We x

have that lim y = 0 (and so y has y = 0 as a horizontal asymptote), but x ( ) sin x 2 that lim x = lim + 2 cos x 2, which does not exist. x x 2 44. Let f(x) = sin x2 x. (a) Compute lim f(x). x (b) Compute lim f (x). (If this limit does not exist, say so.) x (c) Compute lim f (x). (If this limit does not exist, say so.) x This was anticipated in the previous exercise. 45. Let x = t 2 + t and let y = cos t. (a) Find / as a function of t. = dt / dt = sin t 2t +. (b) Find d ( ) as a function of t. dt Using the quotient rule, (c) Find d ( ) d d dt ( ) as a function of t. ( ) = d ( ) sin t dt 2t + (2t + ) cos t + 2 sin t =. (2t + ) 2 = d ( ) / dt dt (2t + ) cos t + 2 sin t = /(2t + ) (2t + ) 2 (2t + ) cos t + 2 sin t = (2t + ) 3

46. Given the equation y 2 + x 2 = xy, compute both / and d 2 y/ 2. Computing is routine; using implicit differentiation one arrives at = y 2x 2y x. Computing d2 y is rather more complicated: 2 d 2 y = (y 2)(2y x) (y 2x)(2y ) 2 (2y x) 2 2(y 2x)2 (y 2x) 2(2y x) + (y 2x) 2y x = (2y x) 2 = (y 2x)(2y x) 2(2y x)2 2(y 2x) 2 + (y 2x)(2y x) (2y x) 3 = 2(3y2 3xy + x 2 ) (2y x) 3. 47. Compute /, given that (a) y = e 2x cos x = 2e2x cos x e 2x sin x = e 2x (2 cos x sin x) ln x (b) y = e = (c) y = ln( + x 2 ) = x + x 2. 48. Consider the curves defined by the equations y = f(x)= 2 x2 + 4 and y = g(x) = ln x. Show that at the point of intersection of these two curves, the tangent lines are perpendicular. (Hint: what is f (x)g (x)? What does this mean?) f (x) = x and g (x) = x. Therefore, at the point of intersection the tangent lines have negative reciprocal slopes.

49. Let y = x x 2 + 3 and compute / using logarithmic differentiation. x + 2 Starting with ln y = ln x + 2 ln(x2 + ) 3 ln(x + 3), we now differentiate both sides and get: and so y y = x + x x 2 + 3(x + 3), y = y x + xy x 2 + x2 + = 3 + x + 2 y 3(x + 3) x 2 x x 2 + x2 + 3 x + 2 (x + 2) 4/3. There s not much point in trying to simplify this further!