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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
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 ( 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.

(a)

To determine

To show: Zero is a critical point of the functions f(x) , g(x) and h(x) but the derivatives f(x) , g(x) and h(x) changes its sign infinitely often on both sides of zero.

Explanation

Given:

The functions are, f(x)=x4sin1x , g(x)=x4(2+sin1x) and h(x)=x4(2+sin1x) .

Proof:

It is known that the critical numbers are exist for which the derivative is zero.

Obtain the derivative of f(x) .

f(x)=ddx(x4sin1x)=x4(sin1x)+(sin1x)(x4)=x4(cos1x)(1x2)+(sin1x)4x3=4x3sin1xx2cos1x

Substitute x=0 in f(x) ,

f(0)=4(0)3sin1(0)(0)2cos1(0)=00=0

Thus, f(x) has a critical point at x=0 .

Obtain the derivative of g(x) .

g(x)=ddxx4(2+sin1x)=x4(2+sin1x)+(2+sin1x)(x4)=x4(0+cos1x(1x2))+(2+sin1x)(4x3)=x2cos1x+8x3+4x3sin1x

Simplify the above as follows.

g(x)=8x3+4x3sin1xx2cos1x=8x3+f(x)

Substitute x=0 in g(x) ,

g(0)=8(0)3+f(0)=0+0=0

Thus, g(x) has a critical point at x=0 .

Obtain the derivative of h(x) .

h(x)=ddxx4(2+sin1x)=x4(2+sin1x)+(2+sin1x)(x4)=x4(0+cos1x(1x2))+(2+sin1x)(4x3)=x2cos1x8x3+4x3sin1x

Simplify the above as follows.

h(x)=8x3+4x3sin1xx2cos1x=8x3+f(x)

Substitute x=0 in h(x) ,

h(0)=8(0)3+f(0)=0+0=0

Thus, h(x) has a critical point at x=0 .

Therefore, it is proved that the functions f(x) , g(x) and h(x) has the critical point at x=0 .

To check the derivative of the functions f(x) changes its sign often on both sides of zero at infinite times.

As the derivative functions involves with trigonometric ratios, consider a number in terms of π .

It is enough to prove that the sign of f(x) changes between any two consecutive terms.

Assume the consecutive terms as x2n=12nπ and x2n+1=1(2n+1)π .

It can be expressed as 1x2n=2nπ and 1x2n+1=(2n+1)π .

The value of sin(2nπ) is always zero ( n is an integer) and cos(2nπ) is always 1 ( n is an integer). Similarly, the value of sin(2n+1)π is always zero and cos(2n+1)π is always 1 . Now find the value of f(x) for the two consecutive terms as follows

(b)

To determine

To show: The function f(x) has neither a local maximum nor a local minimum at zero, g(x) has a local minimum and h(x) has a local maximum.

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