Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 43

f(x)=x^4-2x^3+x+1
f'(x)=4x^3-6x^2+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.
Now to find the critical numbers, solve for x for f'(x)=0.
4x^3-6x^2+1=0
(2x-1)(2x^2-2x-1)=0
2x-1=0
x=1/2
2x^2-2x-1=0
x=(2+-sqrt(4+8))/4
x=(1+-sqrt(3))/2
Now let us evaluate the function at the critical numbers and at the end points of the interval(-1,3) ,
f(-1)=(-1)^4-2(-1)^3+(-1)+1
f(-1)=1+2=3
f(3)=3^4-2*3^3+3+1=31
f(1/2)=(1/2)^4-2*(1/2)^3+(1/2)+1=21/16
f((1+sqrt(3))/2)=3/4
f((1-sqrt(3))/2)=3/4
So, Absolute maximum=31 at x=3
Absolute minimum =3/4 at x=(1+-sqrt(3))/2