Dt 3 dt - University of California, Berkeley

2x + 3 1. If f HxL = ????????????????????? , then f ? HxL =

3x - 1

-12 x - 7 HAL ????????????????????????????????

H3 x - 1L2

-11 HBL ???????????????????????????????

H3 x - 1L2

3 HCL ???????????????????????????????

H3 x - 1L2

f ? HxL =

H3 x - 1L 2 - H2 x + 3L 3 ?????????????????????????????????????????????????????????????????????? ? ??

=

6x - 2 - 6x - 9 ????????????????????????????????????????????????????????

=

H3 x - 1L2

H3 x - 1L2

11 HDL ???????????????????????????????

H3 x - 1L2

12 x + 7 HEL ???????????????????????????????

H3 x - 1L2

-11 ??????????????????????????????? and the answer is B H3 x - 1L2

2. The legs and hypotenuse of the right triangle below are strictly increasing with time. At the instant when a = 6 and

b = 8,

da 1 dc

db

dc

???????? = ????? ???????? and ???????? = k ???????? .

dt 3 dt

dt

dt

What is the value of k at this given instant?

c a

b

1 HAL ?????

4

1 HBL ?????

2

HCL 1

HDL 2

HEL It cannot be determined from the information given.

In the given right triangle, a2 + b2 = c2

da

db

dc

and if we differentiate implicitly, we have 2 a ???????? + 2 b ???????? = 2 c ???????? or

dt

dt

dt

da

db

dc

a ???????? + b ???????? = c ????????

dt

dt

dt

Now we may substitute the given expressions to get

6

ijj

1 ?????

dc ????????

yzz

+

8 ijjk

dc ????????

zzy

=

dc 10 ????????

or

k 3 dt {

k dt {

dt

dc

dc

dc

2 ???????? + 8 k ???????? = 10 ???????? ? 2 + 8 k = 10 so k = 1

dt

dt

dt

And the answer is C

3. Consider the graph of f ? below. If f H5L = -1, then f H0L =

y 2

1

f ?HxL

x 12345 -1

-2

HAL - 1.5

HBL - 1

HCL - 0.5

1 If f H5L = -1, then f H0L = -1 + 1 + ????? - 2

2

HDL 0 = -1.5

HEL 0.5 and the answer is A

4. At x = 1, the function defined by f HxL = "3 #x### + 3,

x?1

1

11

????? x + ???????? , x > 1

is

3

3

HAL undefined

HBL continuous but not differentiable

HCL differentiable but not continuous

HDL neither continuous nor differentiable

HEL both continuous and differentiable

The pieces match up at x = 1, so it is continuous, and

1

1

1

f ? HxL = ?3???????3????!x?!??!?2!?!!?? , x < 1,

and ????? , x > 1 3

and since these match up, f ? H1L = ????? 3

and

the function is both continuous and differentiable at x = 1

and the answer is E

5. The side length of a cube is increasing Hall lengths are increasing at the same rate, so that it remains a cubeL. At a certain instant, the rate of increase in the surface area of the cube is numerically equal to the rate of increase in its volume. What is the length of a side of the cube at that instant?

1 HAL ?????

2

HBL 1

HCL 2

HDL 4

HEL 8

S = 6 s2 and V = s3

so

or 12 s = 3 s2

? 4 s = s2

dS

ds

???????? = 12 s ????????

dt

dt

dV

ds

and ???????? = 3 s2 ????????

dt

dt

and we know that

? s2 - 4 s = 0 ? s ? 4

and the answer is D

ds

ds

12 s ???????? = 3 s2 ????????

dt

dt

6. If f ? HxL = ? x + 2 ?, which of the following could be the graph of f HxL ?

y

y

y

HAL

4

HBL

4

HCL

4

3

y = fHxL 2

1

3

y = fHxL 2

1

y = fHxL 3

2 1

x

-5 -4 -3 -2 -1

1

-1

x

-5 -4 -3 -2 -1

1

-1

x

-5 -4 -3 -2 -1

1

-1

-2

-2

-2

y

y

HDL

4

HEL

4

3

y = fHxL 2

3

y = fHxL

2

1

1

x

-5 -4 -3 -2 -1

1

-1

x

-5 -4 -3 -2 -1

1

-1

-2

-2

?

f ? HxL = -x - 2 on H-?, -2D

so

f HxL

=

-1 ????????

x2

-

2 x

+

C

on

2

x + 2 on H-2, ?L

1 ????? x2 + 2 x + D 2

and the constants need to be such that the two pieces meet up at x = -2, so

f HxL

=

-1 ????????

x2

-

2 x

-

1

on

H-?, -2E

2

1 ????? x2 + 2 x + 3 2

on H-2, ?L

and the answer is E

H-?, -2E on H-2, ?L

7. Evaluate

d ???????? dx

ijjjjjjj2 k1

et2

dtyzzzzzzz {

HAL 0

HBL e

e4 - e HCL ?????????????????????

3

HDL e4 - e

The derivative of any constant is 0

and the answer is A

HEL Cannot be determined from the information given

8. If f HxL = arccot Hx2L, then f ? HxL =

1 HAL ?????????????????????

1 + x4

-2 x HBL ?????????????????????

1 + x4

-1 du

Dx Harccot uL =

????????????????????? ???????? 1 + u2 dx

so

2x HCL ?????????????????????

1 + x4

-4 x3 HDL ?????????????????????

1 + x4

Dx Harccot x2L =

-1 ????????????????????? H2 xL 1 + x4

4 x3 HEL ?????????????????????

1 + x4 and the answer is B

p 9. 4 sin Jx + ????? N =

3

HAL 2 cos x + 2 "#3### sin x HDL 2 sin x + 2 "#3### cos x

HBL 2 cos x - 2 "#3### sin x HEL 2 sin x - 2 "#3### cos x

HCL 2 "#3### sin x - 2 cos x

p

p

p

4 sin Jx + ????? N = 4 sin x cos J ????? N + 4 cos x sin J ????? N

3

3

3

=

4

sin

x

ijj k

1 ????? 2

yzz {

+

4

cos

x

ikjjjjjj

?!!!! 3

???????????? 2

y{zzzzzz

=

2 sin x + 2 "#3### cos x

and the answer is D

dy 10. Let y = f HxL be the solution to the differential equation ???????? = y - 2 x + 1 with the initial condition f H-1L = -2.

dx What is the approximation for f H1L if Euler' s Method is used, starting at x = -1 with 2 steps of equal size?

HAL - 3

HBL - 2

HCL - 1

HDL 0

HEL 1

Our starting point is H-1, -2L, and y1 = -2 + H-2 - 2 H-1L + 1L H1L = -1 Now, once more, with y2 = -1 + H-1 - 0 + 1L H1L = -1 and so f H1L ? -1

and the new point is H0, -1L and the answer is C

11. If the position of a particle is given by the function s HtL = 2 t2 - 3 t + 4 average velocity of the particle on the given interval.

on the interval @1, 5D, then find the

HAL 4

19 HBL ????????

4

HCL 7.2

HDL 9

39 HEL ????????

4

5

1

1

1

36

vave

=

?????????????????? 5-1

v HtL dt

= ????? Hs H5L - s H1LL = ????? H39 - 3L = ???????? =

4

4

4

9

1

and the answer is D

12. The figure below shows the graph of the polynomial function f. For which value of x is it true that f ? ? HxL > f HxL > f ? HxL ?

y

fHxL

x

a

b

c

de

HAL a

HBL b

HCL c

By inspection, the answer is A

HDL d

HEL e

13. If

g is the function given by

?x2???

g HxL = "#t#3#####+######1## dt

1

1 HAL ?????

2

3 HBL ?????

2

9 HCL ?????

2

HDL 8

then g ? H4L = 25

HEL ???????? 2

?x2???

If g HxL = "#t#3#####+######1## dt ,

1

and the answer is B

then

g ? HxL

=

ikjjjjjj$%J%%?%x2?%??%?%N%%3%%%%+%%%%%%1%%%

yzzzzzz {

ijj k

1 ????? 2

yzz {

$%%?4?%?%?3%?%%%%+%%%%%1%%%

8

3

so g ? H4L = ???????????????????????????????? = ?????

2

2

d 14. Find ???????? Hf H2 xL g HxLL at x = 2, given the functions defined graphically below.

dx

y 6

5

gHxL

4

3

2

1

fHxL

x 123456

HAL - 1

-1 HBL ????????

2

1 HCL ?????

2

HDL 7

HEL Does Not Exist

Using the Product Rule, we have

d ???????? Hf H2 xL g HxLL dx

f ? H4L 2 g H2L + f H4L g ? H2L = 0 H2L H4L + 1 H-1L =

= f ? H2 xL 2 g HxL + f H2 xL g ? HxL

-1

and the answer is A

and at

x = 2,

we have

x2 15. What is the slope of the normal line to the curve 4 y + 1 = ??????? + 2 sin y at the point H-2, 0L?

4

HAL - 4

HBL

-1 ????????

2

HCL

-1 ????????

4

HDL 2

HEL 4

2x Using implicit differentiation, we have 4 y ? = ???????? + 2 cos y y ?

4

and plugging in x = -2 and y = 0, we have

4 y ? = -1 + 2 H1L y ?

-1 ? y ? = ???????? so the slope of the normal line is 2, and the answer is D

2

16. If g HxL is a differentiable function such that g H-1L = 3 and g ? H-1L = 4, then which of the following

statements must be true?

I. lim g HxL = lim g HxL

x ? -1+

x ? -1-

g H-1 + hL - 3

II. lim ?????????????????????????????????????????????? = 4

h?0

h

g HxL - 3 III. lim ???????????????????????????? = 4

x ? -1 x + 1

HAL II only

HBL I and II only

HCL I and III only

HDL II and III only

HEL I, II, and III

I is true because the function is continuous at x = -1, and II and III are both true because they are the definitions for

the derivative at x = -1,

and the answer is E

17. The figure below shows the graph of f ?, the derivative of the function f, on the interval @-3, 6D. If the derivative of the function g is given by g ? HxL = 3 f ? HxL, how many points of inflection does the graph of g have on the interval @-3, 6D ?

y 2

1

-3 -2 -1 -1

y = f ?HxL

x 123456

-2

HAL One

HBL Two

HCL Three

HDL Four

HEL Five

We are looking for where the slope of f ? HxL is 0 Hand where f ? ? HxL would change signL. It looks like this is occurring

3 times, at approximately x = -2.1, x = 2.1, and x = 4.7

and the answer is C

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