Answer Key!!! *smile*

[Pages:40]Answer Key!!! *smile* Review Packet for Exam #2

Professor Danielle Benedetto

Math 11

Differentiation Rules Differentiate the following functions. Please do not simplify your derivatives here.

1. y = sin3(x3) y = 3 sin2(x3) ? cos(x3) ? 3x2 = 9x2 sin2(x3) cos(x3)

2. y = cos2(3x) y = 2 cos(3x) ? (- sin(3x)) ? 3 = -6 cos(3x) sin(3x)

3. f (t) = t2 sin5(2t) f (t) = t2[5 sin4(2t) ? cos(2t) ? 2] + sin5(2t) ? 2t = 10t2 sin4(2t) cos(2t) + 2t sin5(2t)

4. H(x) =

1

-

2 x2

5

H(x) = 5

1

-

2 x2

4

?

(4x-3)

=

20 x3

1

-

2 x2

4

5. f (x) = 3 x3 + 8

f (x)

=

1 3

(x3

+

8)-2/3

?

3x2

=

(x3

x2 + 8)2/3

6.

g(t) =

t3 + tan 1 + t2

1 t

(1 + t2) 3t2 + sec2 g(t) =

1 t

?

-

1 t2

-

(1 + t2)2

t3 + tan

1 t

? (2t)

=

t4

+

3t2

-

(1

+

t-2) sec2 (1 + t2)2

1 t

- 2t tan

1 t

7.

p(x) =

1 (-2x + 3)5

p(x)

=

(-2x

+

3)-5,

so

p(x)

=

-5(-2x

+

3)-6

?

(-2)

=

10 (-2x +

3)-6

8.

r(x) =

(2x + 1)3 (3x + 1)4

r(x)

=

(3x

+

1)4

?

3(2x

+

1)2

? (2) - (2x (3x + 1)8

+

1)3

?

4(3x

+

1)3

?

(3)

=

6(2x

+

1)2(3x

+ 1)3[(3x + (3x + 1)8

1)

-

2(2x

+

1)]

=

6(2x

+

1)2[3x + 1 - (3x + 1)5

4x

-

2]

1

=

6(2x + 1)2(-x (3x + 1)5

-

1)

=

-6(x + 1)(2x + (3x + 1)5

1)2

9. S(x) =

1 + 2x 4 1 + 3x

S(x) = 4

1 + 2x 1 + 3x

3

?

(1 + 3x) ? 2 - (1 + 2x) ? 3 (1 + 3x)2

=

4(1

+

2x)3 ? [2 + 6x (1 + 3x)5

-

3

-

6x]

=

-4(1 + 2x)3 (1 + 3x)5

10. g(x) = cos(3x) sin(4x) g(x) = cos(3x) cos(4x) ? 4 + sin(4x) - sin(3x) ? (3) = 4 cos(3x) cos(4x) - 3 sin(3x) sin(4x)

11.

g(x) =

cos(3x) sin(4x)

g(x) = sin(4x)

-

sin(3x) ? 3 - cos(3x) sin2(4x)

cos(4x)

?

4

=

-3

sin(3x) sin(4x) - 4 cos(3x) cos(4x) sin2(4x)

12.

g(x)

=

(x

+

7x-6

) 2x

+

1

cos2(6x)

g(x)

=

(x + 7x-6)

2x

+

1

cos2(6x)

,

so

g(x)

=

(x + 7x-6)

2x + 1 ? 2 cos(6x) ?

2x

+

1 cos2(6x)

(1

-

42x-7)

- sin(6x) ? (6) + cos2(6x) ? 1

?2 +

2 2x + 1

= (x + 7x-6) cos(6x)

-

12 2x

+

1

sin(6x)

+

cos(6x)

+

2x

+

1(1

-

42x-7)

cos2

(6x)

2x + 1

13. w = 5(1 + x2)3

x 2x + 1

x 2x + 1 w =

15(1 + x2)2 ? (2x)

-

5(1 + x2)3

x?

1 2 2x+1

?2

+ 2x + 1 ? 1

x2(2x + 1)

=

30x2(2x

+

1)(1

+ x2)2 - 5(1 + x2)3[x x2(2x + 1)3/2

+

(2x

+

1)]

5(1 + x2)2 6x2(2x + 1) - (3x + 1)(1 + x2)

=

x2(2x + 1)3/2

5(1 + x2) 12x3 + 6x2 - 3x - 1 - 3x3 - x2

=

x2(2x + 1)3/2

=

5(1

+

x2)(9x3 + 5x2 - x2(2x + 1)3/2

3x

-

1)

14. y =

(x2 + 3x)4 + x

-

5 7

y

=

-

5 7

(x2 + 3x)4 + x

-12/7[4(x2 + 3x)3(2x + 3) + 1]

15. g(t) = cos sin3 t t+1

g(t) = - sin sin3 t t+1

? 3 sin2

t t+1

? cos

t t+1

?

t

+

1

?

(1)

-

t

?

1 2 t+1

?

1

t+1

2

Note

that

t + 1 ? (1) - t ? t+1

1 2 t+1

?1

=

t

+

1

-

t 2

(t + 1)3/2

=

t+2 2(t + 1)3/2

.

So

g(t)

=

-3(t 2(t +

+ 2) 1)3/2

sin

sin3

t t+1

sin2 t

cos t

t+1

t+1

16. g(x) = cos2(6x) tan(-x)

2x + 1

g(x) = cos2(6x)

2x

+

1

?

sec2(-x)

?

(-1)

-

tan(-x)

1 2 2x+1

(2)

2x + 1

+ tan(-x) ? 2 cos(6x) ? - sin(6x) ? (6) 2x + 1

= cos2(6x)

-(2x + 1) sec2(-x) - tan(-x) (2x + 1)3/2

- 12 tan(-x) cos(6x) sin(6x) 2x + 1

(Note: This problem could be made slightly easier to solve if you observe that tan(-x) = - tan x before you differentiate.)

Absolute Extreme Values

17. Find the absolute maximum and absolute minimum values of the following functions on the given intervals.

(a)

f (x) = 3x2/3 -

x 4

on

[-1, 1].

f is continuous on [-1, 1], so we use the closed interval method. We compute f (x) =

2x-1/3 - 1/4. On the interval [-1, 1], f is undefined at x = 0, so x = 0 is a critical point

(since f is defined but not differentiable there). Also, f (x) = 0 exactly when x1/3 = 8;

that is, when x = 512; but x = 512 is not in [-1, 1]. The endpoints ?1 also need to be

checked, of course. Testing the critical points and endpoints, we see

f (-1)

=

13 4

,

f (0) = 0,

f (1)

=

11 4

.

So the absolute maximum value is 13/4 (attained at x = -1), and the absolute minimum

value is 0 (attained at x = 0).

(b)

h(x) =

x2 - 1 x2 + 1

on

[-1, 3].

h(x)

=

2x(x2 + 1) - 2x(x2 - 1) (x2 + 1)2

=

(x2

4x +

1)2

.

On the interval [-1, 3], h is always

defined (since it is a rational function and its denominator is never zero). Also, h(x) = 0

happens only when the numerator is zero; that is, when 4x = 0, so x = 0. Applying the

closed interval method:

h(-1) = 0,

h(0) = -1,

h(3)

=

4 5

.

So the absolute maximum value is 4/5 (attained at x = 3), and the absolute minimum

value is -1 (attained at x = 0).

(c)

F (x) = -2x3 + 3x2

on

[-

1 2

,

3].

F (x) = -6x2 + 6x, which is always defined. Setting F = 0 gives -6x(x - 1) = 0, so

x = 0 and x = 1, both of which are in the interval, are the only critical points. Applying

the closed interval method:

3

F

-

1 2

=

1 4

+

3 4

=

1,

F (0) = 0,

F (1) = -2 + 3 = 1,

F (3) = -54 + 27 = -27.

So the absolute maximum is 1, attained at x = -1/2 and x = 1, and the absolute

minimum is -27, attained at x = 3.

2

(d) f (x) = x 3 on [-1, 8].

f (x)

=

2 3

x-1/3

=

3

2 3x

,

which

is

not

defined

at

x

=

0,

but

which

is

defined

everywhere

else. Solving f = 0 gives 2 = 0, which never happens. So the only critical point is

x = 0, which is in the interval. Applying the closed interval method:

f (-1) = 1, f (0) = 0, f (8) = 4.

So the absolute maximum is 4, at x = 8, and the absolute minimum is 0, at x = 0.

(e)

f (x) =

1 1 + x2

on

[-3, 1].

f (x)

=

-(1 + x2)-2

? (2x)

=

(1

-2x + x2)2

,

which

is

always

defined.

(Note

that

the

denom-

inator is never zero.) Solving f = 0 gives 2x = 0, so x = 0, which is in the interval.

Applying the closed interval method:

f (-3)

=

1 10

,

f (0) = 1,

f (1)

=

1 2

.

So the absolute maximum is 1, at x = 0, and the absolute minimum is 1/10, at x = -3.

Curve Sketching For each of the following functions, discuss domain, vertical and horizontal asymptotes, intervals of increase or decrease, local extreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve.

18. f (x) = x3 - 3x2 + 3x + 10

? Domain: f (x) has domain (-, )

? VA: It is a polynomial, continuous everywhere, and so has no vertical asymptotes.

? HA: There are no horizontal asymptotes for this f since lim f (x) = and lim f (x) =

x

x-

-.

? First Derivative Information:

We compute f (x) = 3x2 - 6x + 3 and set it equal to 0 and solve for x to find critical numbers. The critical points occur where f is undefined (never here) or zero. The latter happens when

3x2 - 6x + 3 = 3(x - 1)(x - 1) = 0 = x = 1

As a result, x = 1 is the critical number. Using sign testing/analysis for f ,

f

'

t

E

1

f

or our f chart is 4

x f (x) f (x)

(-, 1)

(1, )

So f is increasing on (-, ). Moreover, f has no extreme values.

? Second Derivative Information:

Meanwhile, f is always defined and continuous, and f = 6x - 6 = 0 only at our possible inflection point x = 1. Using sign testing/analysis for f ,

f

'

t

E

f or our f chart is

1

infl. point

x f (x) f (x)

(-, 1)

(1, )

So f is concave down on (-, 1) and concave up on (1, ), with an inflection point (1, 11) at x = 1.

? Piece the first and second derivative information together:

'

t

1

f

E

f

f

infl. pt.

? Sketch:

5

19.

f (x) =

3x2 1 - x2

? Domain: f (x) has domain {x|x = ?1}

? VA: Vertical asymptotes x = ?1.

? HA: Horizontal asymptote is y = -3 for this f since lim f (x) = -3 because

x?

lim

x?

3x2 1 - x2

?

1 x2

1 x2

=

lim

x?

1 x2

3 -

1

=

-3

? First Derivative Information:

We

compute

f (x) =

6x (1 - x2)2

and

set

it

equal

to

0

and

solve

for

x

to

find

critical

numbers.

The critical points occur where f is undefined or zero. The former happens when x = ?1,

but x = ?1 were not in the domain of the original function, so they aren't technically critical

numbers. The latter happens when x = 0. As a result, x = 0 is the critical number. Using sign testing/analysis for f ,

or our f chart is

f

'

t

t

t

E

-1 0

1

f

local

min

x f (x) f (x)

(-, 0)

(0, )

6

So f is decreasing on (-, -1) and (-1, 0) and increasing on (0, 1) and (1, ). Moreover, f has a local min at x = 0 with f (0) = 0.

? Second Derivative Information:

Meanwhile,

f

=

6(1 + 3x2) (1 - x2)3

.

Using

sign

testing/analysis

for

f ,

f

'

t

t

E

f

-1

1

or our f chart is

x f (x) f (x)

(-, 1)

(-1, 1)

(1, )

So f is concave down on (-, -1) and(1, ) and concave up on (-1, 1).

? Piece the first and second derivative information together:

'

t

t

t

E

-1 0

1

f

f

f

lmocianl

? Sketch:

7

20.

f (x) =

x x-2

? Domain: f (x) has domain {x|x = 2}

? VA: Vertical asymptote at x = 2.

? HA: Horizontal asymptote at y = 1 for this f since lim f (x) = 1 because

x?

lim

x?

x

x -

2

?

1 x

1 x

=

lim

x?

1

1 -

2 x

= 1.

? First Derivative Information:

We

compute

f (x) =

-2 (x - 2)2

to

find

critical

numbers.

The

critical

points

occur

where

f

is

undefined or zero (never here). The former happens when x = 2, which was not in the domain

of the original function. As a result, there are no critical numbers. Using sign testing/analysis for f around x = 2,

f

'

t

E

2

f

or our f chart is

x f (x) f (x)

(-, 2)

(2, )

So f is decreasing on (-, 2) and (2, ) . Moreover, f has no extreme values.

? Second Derivative Information:

8

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