   Chapter 4.4, Problem 38E Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Solutions

Chapter
Section Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

Using the Second Derivative Test InExercises 37-58, find all relative extrema of the function. Use the Second Derivative Test where applicable. f ( x ) = − x 3 + 7 x 2 − 15 x

To determine

To calculate: The relative extrema of the provided function f(x)=x3+7x215x using the second derivative test.

Explanation

Given:

The function f(x)=x3+7x215x.

Formula used:

For a function f that is twice differentiable on an open interval I, if f'(c)=0 for some c, then,

If f''(c)>0 the function f has relative minima at c if f''(c)<0 the function f has relative maxima at c.

Calculation:

First differentiate the provided function,

f'(x)=3x2+14x15

Equate the first derivative to zero to obtain the critical points.

3x2+14x15=03x29x5x+15=0(x3)(3x5)=0x=53,3

Now differentiate the function for the second time and replace this for x:

f''(53)=6(53)+14>0

This implies that the function has relative minima at this point

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