   Chapter 3.3, Problem 77E

Chapter
Section
Textbook Problem

# The three cases in the First Derivative Test cover the situations one commonly encounters but do not exhaust all possibilities. Consider the functions f, g, and h whose values at 0 are all 0 and, for x ≠ 0 , f ( x ) = x 4 sin 1 x    g ( x ) = x 4 ( 2 + sin 1 x ) h ( x ) = x 4 ( − 2 + sin 1 x ) (a) Show that 0 is a critical number of all three functions but their derivatives change sign infinitely often on both sides of 0.(b) Show that f has neither a local maximum nor a local minimum at 0, g has a local minimum, and h has a local maximum.

To determine

a)

To show:

The 0 is a critical number of all three functions but their derivatives change sign infinitely often on both sides of 0

Explanation

1) Concept:

The First Derivative Test-Suppose that c is a critical number of a continuous function f

i. If f' changes from positive to negative at c, then f has a local maximum at c

ii. If f' changes from negative to positive at c, then f has a local minimum at c

iii. If f' is positive to the left and right of c, or negative to the left and right of c, then f has no local maximum or minimum at c

2) Given:

The functions fx=x4sin1x,  gx=x42+sin1x,   hx=x4-2+sin1x

3) Calculation:

Given that

fx=x4sin1x

Differentiate f by product rule also use chain rule

f'x=x4cos1x-1x2+sin1x(4x3)

Simplify

f'x=4x3sin1x-x2cos1x

To find critical number, equate f'x=0

4x3sin1x-x2cos1x=0

Solve for x

Therefore, x=0

Therefore, the critical number is 0

Given that

gx=x42+sin1x

Simplify

gx=2x4+x4sin1x

But fx=x4sin1x

gx=2x4+f(x)

Differentiate g

g'x=8x3+f'(x)

To find critical number, equate g'x=0

8x3+f'(x)=0

Solve for x

Therefore, x=0

Therefore, the critical number is 0

Given that

hx=x4-2+sin1x

Simplify

hx=-2x4+x4sin1x

But fx=x4sin1x

hx=-2x4+f(x)

Differentiate h

h'x=-8x3+f'(x)

To find critical number, equate h'x=0

-8x3+f'(x)=0

Solve for x

Therefore, x=0

Therefore, the critical number is 0

It is given that f0=0

By definition of differentiability

f'0=limx0fx-f(0)x-0

Substitute fx value in f'0

f'0=limx0x4sin1x-0x-0

Simplify

f'0=limx0x3sin1x

Sine function is always bounded by 1. That is -1 sinx 1

Multiplying throughout the inequality by |x3|,

-x3|x3| sin1xx3

Therefore, by squeeze theorem

limx0-x3limx0x3sin1xlimx0x3

But limx0x3=0

Therefore, above equation become

limx0x3sin1x=0

To determine

b)

To show:

i. The function f has neither local maximum nor minimum at 0

ii. The function g has local minimum

iii. The function h has local maximum

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