   Chapter 10, Problem 7T ### 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

# Find the absolute maximum and minimum for f ( x ) = 2 x 3 − 15 x 2 + 3 on the interval [ − 2 , 8 ] .

To determine

To calculate: The absolute maximum and minimum for the function f(x)=2x315x2+3 on interval [2,8].

Explanation

Given Information:

The provided equation is f(x)=2x315x2+3 and the interval is [2,8].

Formula used:

To find absolute maximum and minimum 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).

3. Substitute the end points into f(x).

The maximum value is absolute maximum and the minimum value is absolute minimum.

Calculation:

Consider the provided equation,

f(x)=2x315x2+3

Calculate the first derivative of the above function with respect to x:

f(x)=6x230x=6x(x5)

Set f(x)=0.

6x(x5)=0

Thus, either 6x=0 or (x5)=0.

First consider 6x=0.

Simplify for x,

6x=0x=0

Now, consider (x5)=0.

Solve for x,

x5=0x=5

Thus, the critical values are at x=0 and x=5.

Substitute 0 for x in f(x)=2x315x2+3:

f(x)=2x315x2+3=2(0)315(0)2+3=3

Substitute 5 for x in f(x)=2x315x2+3

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