`h(x) = 5 - x^2, [-3,1]` Find the absolute extrema of the function on the closed interval.

Textbook Question

Chapter 3, 3.1 - Problem 20 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

mathace's profile pic

mathace | (Level 3) Assistant Educator

Posted on

Given: `h(x)=5-x^2, [-3, 1]`

Find the critical values of the function by setting the derivative equal to zero and solving for x. When the derivative is equal to zero the slope of the tangent line to the function would be a horizontal tangent. Relative extrema will occur at the point(s) where the derivative is equal to zero.

`h'(x)=-2x=0`

`x=0`

Plug in the critical x-value(s) and the endpoints of the closed interval interval into the original function h(x).

`h(x)=5-x^2`

`h(-3)=5-(-3)^2=5-9=-4`

`h(0)=5-(0)^2=5-0=5`

`h(1)=5-(1)^2=5-1=4` 

Examine the h(x) values to determine the absolute extrema.

The absolute maximum of the function will occur at the point (0, 5)

The absolute minimum of the function will occur at the point (-3, 4)

We’ve answered 318,919 questions. We can answer yours, too.

Ask a question