11 - Herricks

66

Exam V

Section I

Part A - No Calculators

1.

B

p.97

2

Y = cas (2x)

~ =

2.

C

2cos(2x)(-sin(2x?) . 2

I.

The units on the axes are equal. At (2,2), the slope == l.

As y-coordinates are approaching 8 (from above and below), the

slope lines are flattening out.

For a given value of x, the slopes vary at different heights.

III.

C

Inrx

=

Then

True

False

Qy =

dx

At (e ,1),

1

-Inx

In x1/2

=

2

C

True

p.98

Y

4.

-4sin(2x)cos(2x)

p.97

II.

3.

=

2

1.1

2

Qy

dx

x

=

1

2e2

p.98

f(x)

=

I x2 - 11

is not differentiable

at x

=

¡À

l.

The graph has sharp corners there.

f(x) = ~

f(x)

is not continuous for x between -1 and 1 because it is undefined there.

=~

is differentiable

for all real

x; f '(x)

= ~.

2

Hence the function is

x

+1

continuous for all real x.

f(x)

=

1

x -1

-2--

is not continuous at x

=¡À1

The third function (C) is both continuous

5.

o

since the function is undefined there.

and differentiable.

p.98

The slope of the line though

Since y

= rx,

we have

(9,3) and (1,1) is m

v' =

1

3-1

9-1

4¡¤

1

2rx¡¤

Since the tangent line is fa have slope m, we have

1

-2rx

Copyright 2006 Venture Publishing

1

-- -4

=> x

=

4.

¡¤ '() Calculators

6.

A

SOLUTIONS

Exam V Section I Part A

-

67

p.99

X

4

5

x

10

f(x) = 2

f '(x)

(x

= 2x 3 - 2"1 x4.

f"(x) = 6x2 - 2x3 = 2x2(3 - x).

r'(x), take

To maximize

f '(x) is an upside-down

Critical numbers for f '(x) are the

quartic, so has a maximum.

zeros of ["(x); they are x = 0 and x = 3. f '(x) is increasing on either side of x = 0, so

that is NOT the maximum. f"(x) >0 when x < 3 and f"(x) < 0 when x > 3. Therefore, f

attains its maximum at x = 3.

E

p.99

=

aCt)

=

v(O) =

vet)

set)

=

4 - 6t

f aCt)dt

20 => C = 20 =>

f vet) dt =

s(3) - s(l)

8.

D

2

4t - 3t + C

=

=

+ 20

(51 + D) - (21 + D)

30

"fX+3 -

lim [fX+3 - 2][fX+3 + 2]

x.....?

1

(l-x)[fX+3

+ 2]

lim

(x + 3) -4

X""" 1 (1- x)[fX+3

+ 2]

lim

x-I

x.....?

l (1- x)[fX+3 + 2]

lim

-1

1

-0.25

2

1- x

1 ..Jx+3

+ 2

=

-4"

p.l00

X"""

at x

= 1(the only place in question), we need

1

But

lim f(x) = lim

X"""

1

Therefore

A

3i

2t2 - t + 20t + D

To achieve continuity

lim f(x) = f(I).

10.

4t -

(18 - 27 + 60 + D) - (2 - 1 + 20 + D)

X"""

A

=

p.99

lim

x.....?

l

9.

vet)

3

2

x

x.....?

l

-2x + 1

=

x-1

lim (x -1)

1

=

0 while f(l)

= k.

X"""

k = O.

p.l00

b

The average value of f(x) on [a,

b] is defined to be

f

b~a

a

1/2

M

=

-1

2" - 0

f (e

2x + 1) dx

_1-.

0

Copyright

2006 Venture Publishing

f(x) dx.

Therefore

68

11.

SOLUTIONS

B

-

Exam V Section I Part A

Multiple-Choice

p. 100

e

t

v(t)

= t

v '()t

=

t

a

If

t

te-e

-2-

t

=

t

itical num ber

,so th e on 1y critic

er iis t

e(t-l)

2

t

= 1.

< t < I, then v'(t) < 0, so v is decreasing.

If t > 1, then v'(t) > a, so v is increasing.

Thus v achieves its minimum value at t = 1.

12.

D

p.l0l

f~dX

- 2f ~l+x

l+x

(Note that

13.

D

l+x

2

dx

= 1

= a

f "(1) = a

.Thus a = -6;

J

f'(O)

b

1

=

2

x2_x

--dx3

x

1

f

(x-l_x-2)

dx

=

[lnlxl +

1

1.

~r

1

=

C

=b

a=a

a = 12 + 2a

~

~

~

f(O)

p.101

2

15.

+C

p.l0l

= x4 + ax 2 + b

f'(x) = 4x3 + 2ax

r" (x) = 12x 2 + 2a

A

2In(1 + /)

is always positive-valued.)

f(x)

14.

=

(In 2 +

1

p.l02

Let x denote the edge of the cube. Then

= x3

dV

crt = 3x2dxdt

SA = 6/

V

We are given that

dx

dt

=

When SA = 150, then x = 5.

dV

0.2. Hence

= 3¡¤5 2 . (0.2) = 15

Copyright

1

2) - (In 1 + 1) = In2 - 2

crt

2006 Venture Publishing

No Calculators

16.

B

SOLUTIONS

-

Exam V Section I Part A

69

p.102

13

o

J

13

xdx

=

~

J 2-flJ

.

2x

-{3

-V 1 + x2 ]

dx

= 2 - 1 = 1

()

o

This can also be done with a formal substitution.

2

1

Let u = x + 1, so that du = 2x dx, and x dx = 2" duo

To change the limits of integration,

4

J

4

~dU

Then

1

17.

D

-

1/2

2

1

U

=>

=>

we note that x = 0

and x = {3

f

u

-1/2d

u

u1/2 ]

U= 1

u = 4.

4

=2-1=1

1

p.102

Solution I.

Work

analytically.

If f(x) = In

,

then f (x)

I x2 - 4 I

on the interval

2x

= / _4

(-2,2),

2x

= (x - 2)(x + 2) .

(A)

(B)

f'(x) < 0 when 0 < x < 2; f is decreasing then.

f(O) = In 4 :t: 0, so (0,0) is not on the graph.

(C)

1x2 - 41 has a maximum value of 4 on the given

False.

False.

domain, so f(x) has a maximum value of In(4).

Since f(O) = In 4, f does not have an asymptote

at x = O.

Thus (D) must be True.

False.

(E)

Solution

II.

Work

graphically.

The graph of y =

f(x) = In(y) = In

I/ - 4 I

then has

the darker graph shown. The

selection of the correct answer is

made easier with these graphs.

18.

E

p.103

g(x) = Arcsin(2x)

,

g (x)

=

1

--J 1 - (2x)2

¡¤2

y=: , I x2-41

I x2 - 4 I ,

restricted to the open interval

(-2,2), is the tip of an upside-down

parabola, as shown to the right.

2

Copyright 2006 Venture Publishing

False.

,

,

,

,

,

-2',

,

,

,

,

,,

,,

1

,,

'2

,

: y = 1n

I x2-

4

I

70

19.

SOLUTIONS

B

x(x

2

4

21

-1) dx

f

B

2

-1) 4 (2xdx)

(x -1)

5

5

+C

II.

f(2)

=

o

=

1; f' (1)

1.

False

1

f f(x) dx

= -1

?'(3.5)

= -2 ;

1

2

f f(x) dx

=

f f(x) dx

-1 ;

=

-1

-1

True

i

False

p.104

g(x)

g'(x)

= {X (x _1)2/3

= 2~(x _1)2/3

+ ~ (x _1)-1/3

x must be positive (so that

{X

1,- exists).

2'1 x

1 (so that (x _lr1/3 exists).

x must not equal

The domain is { x

D

+C

kx

III.

23.

5

ke

1

C

2

(x -1)

p.l04

I.

22.

1

10

=

p.l03

~

dx --

21.

(x

2

1

2¡¤

B

Multiple-Choice

p. 103

f

20.

Exam V Section I Part A

-

I

x > 0 and x ::F 1 } = { x

I

0 < x < 1 or x > 1 } .

p.104

x(t) = lOt - 4t2

x'(t) = 10 - 8t

x'(t) > 0 if t <

t. The point

is moving to the right.

x'(t) < 0 if t > ~. The point is moving to the left.

The total distance

Since

x(

?)

= ~ '

traveled

x(l)

=

is [x(

t) - X(I)] + [x( ~) - X(2)].

6, and x(2)

Copyright

=

4, the total distance traveled is

2006 Venture Publishing

~.

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