Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 4

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