BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 1RE
To determine

To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function f(x)=x39x2+24x2 on the interval [0, 5].

Expert Solution

Answer to Problem 1RE

The local maximum occurs at x=2 .

The local minimum occurs at x=4 .

The absolute maxima occurs at x=2,5 .

The absolute minimum is x=0 .

Explanation of Solution

Given:

The function is, f(x)=x39x2+24x2 and the interval is, [0,5] .

Calculation:

Obtain the first derivative of the given function.

f'(x)=3x218x+24

Set f(x)=0 and obtain the critical numbers.

3x218x+24=03(x4)(x2)=0x=2,4

The critical numbers are 2 and 4 which lies on the given interval [0,5] .

Apply the extreme values of the given interval and the critical numbers in f(x) .

Substitute x=0 in f(x) ,

f(0)=039(0)2+24(0)2=2

Substitute x=2 in f(x) ,

f(2)=239(2)2+24(2)2=89(4)+482=18

Substitute x=4 in f(x) ,

f(4)=439(4)2+24(4)2=649(16)+962=14

Substitute x=5 in f(x) ,

f(5)=539(5)2+24(5)2=1259(25)+1202=18

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of f(x) is 18 and the absolute minimum of f(x) is −2.

Therefore, the absolute maxima of f(x) occurs at x=2,5 and the absolute minimum of f(x) is occurs at x=0 .

To find the local maximum and minimum obtain the second derivative of the function.

f''(x)=ddx(3x218x+24)=6x18

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute x=2 in f''(x) ,

f''(2)=6(2)18=1218=6<0

Substitute x=4 in f''(x) ,

f''(4)=6(4)18=2418=6>0

Thus, the local maximum occurs at x=2 and the local minimum occurs at x=4 .

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