   Chapter 8.4, Problem 6SWU ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# In Exercises 5 and 6, find all relative extrema of the function. f ( x ) = 1 3 x 3 − 9 x + 2

To determine

To calculate: The all the values of relative extrema of the provided function

f(x)=13x39x+2.

Explanation

Given information:

The provided function is f(x)=13x39x+2.

Formula used:

General power derivative rule:

ddxxn=nxn1

Condition for critical points:

dydx=f(x)=0

If function f(x) is increasing, then f(x)>0 and if function is decreasing, then f(x)<0.

From first derivative test, if f(x) changes positive to negative, then function f(x) has relative maxima and, if f(x) changes negative to positive then function f(x) has relative minima.

In other words, if function f(x) changes from increasing function to decreasing at any point critical, then function f(x) will have relative maxima at that point, and if function f(x) changes from decreasing function to increasing at any critical point, then function f(x) will have relative minima at that point.

Calculation:

Consider the provided function:

f(x)=13x39x+2

Differentiate the above function with respect to x using power derivative rule,

f(x)=ddx[13x39x+2]=13ddx[x3]9ddx[x]+ddx=x29

Apply f(x)=0 in order to find critical points.

x29=0x2=9x=±3

Thus, critical points are x=3, and x=3

Now, rewrite the first derivative of the function.

f(x)=x29

At x=5.

f(5)=(5)29=16=+ve>0

At x=2.

f(2)=(2)29=5=ve<0

At x=5

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