   Chapter 3.3, Problem 16E

Chapter
Section
Textbook Problem

# 15-17 Find the local maximum and minimum values of f using both the First and Second Derivative Tests. Which method do you prefer? f ( x ) = x 2 x − 1

To determine

To find:

The local maximum and minimum values of f using both first and second derivative tests

Explanation

1) Concept:

By using first and second derivative tests

2) Tests:

a) 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' 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

b) Second derivative test:

Suppose f" is continuous near c

i.If f'c=0 and f"(c)>o, then f has local minimum at c

ii. If f'c=0 and f"(c)<o, then f has local maximum at c

3) Formula:

i. Quotient rule:

ddxfxgx= gxddxfx-fxddx(gx)gx2

ii. Difference rule:

ddxfx-gx=ddxfx-ddx(gx)

iii. Constant function rule:

ddxC=0

iv. Power rule:

ddxxn=nxn-1

4) Given:

fx=x2x-1

5) Calculation:

Differentiate f with respect to x,

f'x=ddxx2x-1

By using quotient rule,

f'x=x-1ddxx2-x2ddx(x-1)x-12

By using power rule and difference rule,

f'x=x-1(2x)-x2ddxx-ddx1x-12

By using constant function rule,

f'x=x-12x-x2(1-0)x-12=2x2-2x-x2x-12=x2-2xx-12

Now, find where f' is 0

f'x=x2-2xx-12=0x(x-2)=0x=0 and 2 these are critical numbers

And at x=1f'x is not defined

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