Calculus exam Solutions

8/11/2019 Calculus exam Solutions

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AP Calculus Chapter 3 Testbank

(Mr. Surowski)

Part I. Multiple-Choice Questions (5 points each; please circle the correctanswer.)

1. Iff(x) = 10x4

3 +x, thenf(8) =

(A) 11 (B)41

3 (C)

83

3 (D)

21

3 (E) 21

2. Ifg(x) =3x

2

+x3x2 x , theng(x) =

(A) 1 (B)6x2 + 1

6x2 1 (C) 6

(3x 1)2 (D) 2x2(x2 x)2 (E)

36x2 2x(x2 x)2

3. Iff(x) =

1 +

x, thenf(x) =

(A) 14

x

1 +

x

(B) 1

2

x

1 +

x

(C) 1

4

1 +

x

(D) 1

4x1 +x(E)

12

x

1 +

x


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4. Find dy

dxgiven thatx3y+xy3 = 10.

(A) 3x2 + 3xy2

(B)(3x2 + 3xy2)

(C) 3x2y+y33xy2 +x3

(D)3x2y+y3

3xy2 +x3

(E)x2y+y3

xy2 +x3

5. Iff(x) = sin2 x, findf(x).(A) sin2 x (B) 2 cos 2x (C) cos 2x (D)4sin2x (E) sin2x

6. Iff(x) = 3x, findf(x) =

(A) 3x

ln 3 (B)

3x

ln 3 (C)

3x

(D)

3x1

(E) ln3(3x)

7. Find the slope of the normal line to the graph ofy =x + cos xy at the point(0, 1).

(A) 1 (B) 1 (C) 0 (D) 2 (E) Undefined

8. Iff(x) = 3x2 x andg(x) =f1(x), theng(10) could be

(A) 59 (B) 1

59 (C) 11 (D)

1

11 (E)

1

10

Note that f(x) = 3x2

x =

x = g(3x2

x). Differentiate this and get

1 = (6x 1)g(3x2 x). One of the solutions of 3x2 x= 10 isx = 2; for thisvalue get 1 = 11g(10).

9. If the functionf(x) is differentiable andf(x) =

ax3 6x if x 1bx2 + 4 if x >1,

then

a=

(A) 0 (B) 1 (C) 14 (D)24 (E) 26


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10. Two particles leave the origin at the same time and move along the y-axiswith their respective positions determined by the functions y1 = cos2t andy2 = 4 sin t for 0< t


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14. What is the equation of the line tangent to the graph ofy= sin2 xatx=

4?

(A)y 12

=

x 4

(B) y 1

2=

x

4

(C)y 12

=

x 4

(D)y 12

=1

2

x

4

(E)y 12

=1

2

x

4

15. If the function f(x) =

3ax2 + 2bx+ 1 if x 1ax4 4bx2 3x ifx >1 is differentiable for all

real values ofx, thenb=

(A)114

(B) 1

4 (C)7

16 (D) 0 (E)1

4

16. The position of a particle moving along thex-axis at timet is given byx(t) =ecos2t, 0 t . For which of the following values oftwillx(t) = 0?I.t= 0 II.t=

2 III. t=

(A) I only

(B) II only

(C) I and III only

(D) I and II only(E) I, II, and III


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17. Iff(x) = (3x)3x, thenf(x) =

(A) (3x)3x(3 ln(3x) + 3)

(B) (3x)3x(3 ln(3x) + 3x)

(C) (9x)3x

(ln(3x) + 1)(D) (3x)3x1(3x)

(E) (3x)3x1(9x)

18. Given thatf(x) = 2x2 + 4, which of the following will calculate the derivativeoff(x)?

(A) [2(x+ x)2 + 4] (2x2 + 4)

x

(B) limx0

(2x2 + 4 + x) (2x2 + 4)x

(C) limx0

[2(x+ x)2 + 4] (2x2 + 4)x

(D) (2x2 + 4 + x) (2x2 + 4)

x(E) None of the above.

19. Given that g(x) = 1

x+ 1, which of the following will calculate the derivative

ofg(x)?

(A) 1

x

1

x+ x+ 1 1

x+ 1

(B) lim

x0

1

x 1

x+ x+ 1

1

x+ 1

(C)

limx0

1

x

limx0

1

x+ x+ 1 1

x+ 1

(D) limx0

1

x+ x+ 1 1

x+ 1

(E) None of the above.


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The next two questions pertain to the function f, whose graph is given below:

20. For the functionf,

I.f(

3)> 0

II. f(0)< 0

III. f is differentiableon the interval(0, 1)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I, II, and III

-6 -4 -2 2 4 6

-10

-5

5

10

x

y

y=f(x)

21. For the functionf

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 (1, 5)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) II and III


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22. Given the graph of the rational function fbelow, give a sketch of the graphofy = f(x) on the same coordinate axes. (Note: the graph ofy =f(x)has a vertical asymptote at x= 1.) The graph ofy=f(x) is in blue .

4 -2 2 4

-4

-2

2

4

68

X

Y

23. The following graph represents a functiong defined on the interval [4, 4] anddifferentiable on (4, 4). On the same coordinate axes, graphy =g(x) overthe interval (3, 3).The graph ofy=g (x) is in blue .

4 -2 2 4

-4

-2

2

4

X

Y


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24. The following graph is that ofy = h(x). On the same coordinate axes, givea sketch ofy = h(x), assuming that h(0) = 1. The graph ofy = h(x) is inblue .

4 -2 2 4

-2

-1

1

2

3

4

X

Y

25. The following graph is that ofy = h(x). On the same coordinate axes, givea sketch ofy = h(x), assuming that h(0) = 0. The graph ofy = h(x) is in

blue .

X

Y


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26. Using the definition of the derivative of a function, find f(x), where f(x) =x x4. Then findf(1).

f(x) = limh0

f(x+h) f(x)h

= limh0

(x+h) (x+h)4 (x x4)h

= limh0

(x+h) (x4 + 4x3h+ 6x2h2 + 4xh3 +h4) (x x4)h

= limh0

h (4x3h+ 6x2h2 + 4xh3 +h4)h

= limh0

[1 (4x3 + 6x2h+ 4xh2 +h3)] = 1 4x3.

That is to say, f(x) = 1 4x3, and sof(1) = 1 4 13 = 3.

27. Using the definition of the derivative of a function, find dy

dx where y =

x.

Then find dy

dx

x=4

.

dydx = limh0 f(x+h) f(x)h= lim

h0

x+h x

h

= limh0

(

x+h x)(x+h x)h(

x+h+

x)

= limh0

x+h xh(

x+h+

x)

= limh0

hh(

x+h+

x)

= limh0

1x+h+

x

= 1

2

x

Therefore, dy

dx

x=4

= 1

2

4=

1

4.


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28. Letf(x) = 4x3 21x2 24x+ 23.(a) Computef(x). f(x) = 12x2 42x 24.

(b) Find all values ofx satisfying f(x) = 0. f(x) = 0 12x2 42x 24 =0 2x2 7x 4 = 0 (2x+ 1)(x 4) = 0 x=

12, 4.

29. Letf(x) =x+1

x.

(a) Computef(x).f(x) = 1 1x2

(b) Find all values ofxsatisfyingf(x) = 0.

We have 0 =f(x) = 1 1x2

= x2 1x2 x= 1.

30. Lety= x

1 +x2.

(a) Compute dy

dx.

Using the quotient rule, one has

dy

dx =

ddx(x)(1 +x

2) x ddx(1 +x2)(1 +x2)2

= (1 +x2) x(2x)

(1 +x2)2 =

1 x2(1 +x2)2

.

(b) Compute all values ofxfor which dydx = 0.

From the above, its clear that dy

dx= 0 x= 1.


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31. Lets(x) =sin x

x and compute lim

xs(x).

We have, using the quotient rule, thats(x) =x cos x sin x

x2 . Therefore,

limx s(x) = limx

cos xx sin xx2

= 0 0 = 0.

32. The graph below depicts the velocity v = s(t) of a particle moving along astraight line, where on this straight line positive direction is to the right.

2 4 6 8 10

-4

-2

2

4

t (time)

v (velocity)

(a) Would you say that at time t = 1 the particle is moving to the left,moving to the right, or not moving at all? Please explain.Particleis moving to the right, as v(1)> 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.Particleis 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.Particleis not moving, as v(4) = 0.

(d) Find (estimate) two values oft at which time the particle is not acceler-ating. It appears thata(t) =v (t) = 0 where t 1 or 3. At such valuesoftthe particle will not be accelerating.


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(e) Find (estimate) a value of t at which time the particle is moving to theleft, but withzero acceleration.This would happen att 3 asa(3) 0andv(3)< 0.

(f) According to this graph, at how many distinct times is the particle atrest?The particle is at rest when its not moving; this happens for FIVEvalues oft.

(g) For which values oftis the particle not only at rest, but is not accelerating(i.e., has no forces acting on it)?There are NO SUCH values oft.

(h) According to this graph, at how many distinct times is the particle notaccelerating?We havea(t) = 0 for FOUR values oft.

33. Using logarithmic differentiation compute f(x) where f(x) = xx, x > 0.Starting with ln f(x) = ln xx = x ln x, and differentiating both sides, we getf(x)f(x)

= ln x+ 1, and so f(x) =f(x)(ln x+ 1) =xx(ln x+ 1).

34. LetP(t) = 1

1 +ekt, wherek is a real number.

(a) Show that dP

dt

=kP(1

P).

Using the chain rule, we have

dP

dt =

kekt(1 +ekt)2

= k

1 +ekt e

kt

1 +ekt= kP(1 P).

(b) Show that d2P

dt2= 0 whenP = 1/2.

Using implicit differentiation, together with the result of part (a), we havethat

d2P

dP2=kP(1 P) kP P=kP(1 2P).

From the above, its now obvious that d2P

dP2= 0 when P =

1

2.


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35. Let f and g be differentiable functions and assume that f(1) = 2, f(1) =1, g(1) = 1, g(1) = 0. Compute h(1), given thath(x) =x2f(x)g(x).This uses only the product rule:

h(x) = 2xf(x)g(x) +x2

f(x)g(x) +x2

f(x)g(x).Substituting x= 1 yields

h(1) = 2f(1)g(1) + f(1)g(1) + f(1)g(1) = 2 2 (1 ) + 1 (1 ) + 2 0 = 5.

36. In your text1 it was given as a exercise that the dollar cost of producing x

washing machines is c(x) = 2000 + 100x 0.1x2

. Why is this an absolutelyrediculous cost model? (What is lim

xc(x)? Is this reasonable?)

The above model says that the cost of producingxwashing machines eventu-ally becomes NEGATIVE. This is clearly preposterous!

37. The volume V is a sphere of radius r is given by the formula V = (4/3)r3.Suppose that you know that the radius r is an increasing function of t, and

that whenr= 3,

dr

dt = 2. Compute

dV

dt whenr= 3.

Using the chain rule, dV

dt = 4r2

dr

dtr=3= 4 32 2 = 72.

38. Compute d

dx

cos x

1 + sin x

, simplifying as much as possible.

Using the quotient rule,

d

dx

cos x

1 + sin x

= sin x(1 + sin x) cos2 x

(1 + sin x)2 =

(1 + sin x)(1 + sin x)2

= 1

1 + sin x.

1Exercise 10, page 130.


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39. Note that the point (1, 2) is on the curve defined by y3 xy2 x2y 2 = 0.

(a) Compute dy

dx at the point (1, 2). Using implicit differentiation, one has

3y2y y2 2xyy 2xy x2y= 0, and so

y=dy

dx=

y2 + 2xy

3y2 2xy x2 ,

and so

y(1, 2) = 22 + 2 1 2

3 22 2 1 2 12 =8

7

(b) Find an equation of the straight line tangent to the above curve at thepoint (1, 2). Such an equation can be written as

y 2 =y (1, 2)(x 1),which becomes y 2 = 8

7(x 1).

(c) Find an equation of the straight line normal to the above curve at thepoint (1, 2). Such an equation can be written as

y 2 = 1

y(1, 2)

(x 1),

which becomes y 2 = 78

(x 1).

40. Suppose that y=ex cos x. Show that y+ 2y+ 2y= 0

This is routine: y= ex cos x ex sin x, and soy =ex cos x + ex sin x +e

x

sin x ex

cos x= 2ex

sin x. Therefore,

y+ 2y+ 2y = 2ex sin x+ 2(ex cos x ex sin x) + 2(ex cos x) = 0.


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41. Find f(x), given that f(x) = xx2 + 9

and simplify your result as much as

possible.

Using the quotient and chain rules,

f(x) = x2 + 9 2x2(x2 + 9)1/2x2 + 9

= 9 x2(x2 + 9)3/2

.

42. Computef(1), given thatf(x) = sin

x2 + 3

.

f(x) =x cos

x2 + 3

x2 + 3

; therefore,f(1) = 1/2.

43. Letx= t2 1, y= t 1t+ 1

givexandy parametrically in terms oft.

(a) Compute dy

dx in terms oft.

dy

dx=

dy

dt/

dx

dt =

2

(t+ 1)2 2t= 4t

(t+ 1)2.

(b) Find all values oftfor which dy

dxfails to exist.

From part (a) dydx

fails to exist when t= 1.

(c) Find all values oftfor which dy

dx= 0.

From part (a) dy

dx= 0 whent= 0.

(d) Compute limt

x, limt

y, limt

dy

dx.

limt = +; limt y = 1; limtdy

dx = 0.(e) Suppose that the graph ofy =f(x), where f is a differentiable function,

has a horizontal asymptote (say with limx

y =c, for some real number c.

Would you expect that limx

dy

dx= 0.

This is a bit subtle, but we cannot infer that limx

dy

dx = 0 (even though

we might expect this to happen!). A counterexample isy = sin(x2)

x . We


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have that limx

y= 0 (and so y hasy = 0 as a horizontal asymptote), but

that limx

dy

dx = lim

x

sin x2x2

+ 2 cos x2

, which does not exist.

44. Letf(x) =

sin x2

x .

(a) Compute limx

f(x).

(b) Compute limx

f(x). (If this limit does not exist, say so.)

(c) Compute limx

|f(x)|. (If this limit does not exist, say so.)This was anticipated in the previous exercise.

45. Letx= t2 +t and let y= cos t.

(a) Finddy/dxas a function oft. dy

dx=

dy

dt/

dx

dt = sin t

2t+ 1.

(b) Find d

dt

dy

dx

as a function oft.

Using the quotient rule,

d

dtdy

dx = d

dt sin t

2t+ 1

= (2t+ 1) cos t+ 2 sin t

(2t+ 1)2 .

(c) Find d

dx

dy

dx

as a function oft.

d

dxdy

dx = d

dtdy

dx /dx

dt= (2t+ 1) cos t+ 2 sin t

(2t+ 1)2 /(2t+ 1)

= (2t+ 1) cos t+ 2 sin t

(2t+ 1)3


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46. Given the equation y2 +x2 =xy, compute bothdy/dxandd2y/dx2.

Computing dy

dx is routine; using implicit differentiation one arrives at

dy

dx=

y 2x2y x

.

Computing d2y

dx2 is rather more complicated:

d2y

dx2 =

(y 2)(2y x) (y 2x)(2y 1)(2y x)2

=

(y

2x)

2(2y

x)

2(y 2x)2

2y x + (y

2x)

(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+x2)

(2y x)3 .

47. Computedy/dx, given that

(a) y=e2x cos x dy

dx= 2e2x cos x e2x sin x= e2x(2cos x sin x)

(b) y=elnx dy

dx= 1

(c) y= ln(

1 +x2) dy

dx=

x

1 +x2.

48. Consider the curves defined by the equations y = f(x)= 12

x2 + 4 and y =

g(x) = ln x. Show that at the point of intersection of these two curves, thetangent lines are perpendicular. (Hint: what is f(x)g(x)? What does thismean?)

f(x) = xandg(x) =1x

. Therefore, at the point of intersection the tangent

lines have negative reciprocal slopes.


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49. Lety=x

x2 + 13

x+ 2and computedy/dxusing logarithmic differentiation.

Starting with ln y= ln x +12ln(x2 + 1) 1

3ln(x + 3), we now differentiate both

sides and get:

y

y =

1

x+

x

x2 + 1 1

3(x+ 3),

and so

y = y

x+

xy

x2 + 1 y

3(x+ 3)

=

x2 + 1

3x+ 2 + x2

x2 + 1 3x+ 2 x

x2 + 1

(x+ 2)4/3 .

Theres not much point in trying to simplify this further!

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Calculus exam Solutions