`f(x) = x^3 - 6x^2 + 5, [-3, 5]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 50 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

2 Answers | Add Yours

mathace's profile pic

mathace | (Level 3) Assistant Educator

Posted on

Given: `f(x)=x^3-6x^2+5,[-3,5].`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=3x^2-12x=0`

`3x(x-4)=0`

`x=0,x=4` 

The critical numbers are x=0 and x=4 . Plug in the critical numbers and the endpoints of the interval [-3, 5] into the original f(x) function. 

f(-3)=-76

f(0)=5

f(4)=-27

f(5)=-20

Examine the f(x) values to determine the absolute maximum and absolute minimum values. 

The absolute maximum occurs at the point (0, 5).

The absolute minimum occurs at the point (-3, -76).

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`f(x)=x^3-6x^2+5`

differentiating.

`f'(x)=3x^2-12x`

`f'(x)=3x(x-4)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`3x(x-4)=0`

`x=0 , x=4`

Now evaluate the function at the critical numbers and at the end points of the interval.

`f(0)=0^3-x(0)^2+5=5`

`f(4)=4^3-6(4)^2+5=-27`

`f(-3)=(-3)^3-6(-3)^2+5=-76`

`f(5)=5^3-6(5)^2+5=-20`

So the function has absolute maximum=5 at x=0 

There in no absolute minimum which is clear from the graph.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

1 reply Hide Replies

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

Pl. note correction in absolute minimum.

Function has absolute minimum=-76 at x=-3

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question