   Chapter 10, Problem 6T ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 4-6, use the function y = 3 x 5 − 5 x 3 + 2 .Find the relative maxima and minima of this function.

To determine

To calculate: The relative maxima and minima for the function y=3x55x3+2.

Explanation

Given Information:

The provided equation is y=3x55x3+2.

Formula used:

To find relative maxima and minima of a function,

1. Set the first derivative of the function to zero, f'(x)=0, to find the critical values of the function.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Calculate the sign of the function to the left and right of the critical values.

(a) If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

(b) If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the provided equation,

y=3x55x3+2

Calculate the first derivative of the given equation.

y=15x415x2=15x2(x21)=15x2(x1)(x+1)

Set y=0.

15x2(x1)(x+1)=0

Thus, either 15x2=0 or (x1)=0 or (x+1)=0.

First consider 15x2=0.

Simplify for x.

15x2=0x2=0x=0

Now, consider (x1)=0.

Solve for x.

x1=0x=1

Now, consider (x+1)=0.

Solve for x.

x+1=0x=1

Thus, the critical values are at x=0, x=1 and x=1.

Substitute 0 for x in y=3x55x3+2.

y=3x55x3+2=3(0)55(0)3+2=2

Substitute 1 for x in y=3x55x3+2.

y=3x55x3+2=3(1)55(1)3+2=35+2=0

Substitute 1 for x in y=3x55x3+2.

y=3x55x3+2=3(1)55(1)3+2=3+5+2=4

First consider x=1.

Calculate y' at x=2.

y=15x2(x1)(x+1)=15(2)2(21)(2+1)=15(4)(3)(1)=180

Thus, y>0 on left of x=1.

Calculate y at x=12:

y=15x2(x1)(x+1)=15(12)2(121)(12+1)=15(14)(32)(12)=15(14)(32)(12)

Thus, y<0 on right of x=1

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