# `f(x) = sqrt(x^2 + x + 1), [-2,1]` Find the local and absolute extreme values of the function on the given interval.

### Textbook Question

Chapter 4, Review - Problem 4 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=sqrt(x^2+x+1)`

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.

`f'(x)=1/2(x^2+x+1)^(1/2-1)(2x+1)`

`f'(x)=(2x+1)/(2sqrt(x^2+x+1))`

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

`(2x+1)/(2sqrt(x^2+x+1))=0`

`2x+1=0`

`2x=-1 , x=-1/2`

`f(-2)=sqrt((-2)^2+(-2)+1)=sqrt(3)`

`f(1)=sqrt(1^2+1+1)=sqrt(3)`

`f(-1/2)=sqrt((-1/2)^2-1/2+1)=sqrt(3/4)=sqrt(3)/2`

Absolute maximum= `sqrt(3)` , at x=-2

Absolute maximum =`sqrt(3)` , at x=1

Absolute Minimum = `sqrt(3)/2` , at x=-1/2