Thread: Absoulute max and min

the Absoulute max and min f(x)=x^4 -2x^2 +3 on [-2,3] is at x=?

I do not know how to do this.

here are my thoughts.

1) take derivative of f(x) and set equal to 0

f'(x)= 4x^3 -4x = 0

4x(4x^2-1) =0

x=0

x= +1/2
x= -1/2

f(0) = 3
f(1/2) = 2.54 ???

I am confused

Originally Posted by NeoSonata

f'(x)= 4x^3 -4x = 0
4x(4x^2-1) =0

so
f(0)=3
f(-1)=2
f(1)=2

What am I suppose to find next?

Originally Posted by NeoSonata

so f(0)=3 f(-1)=2 f(1)=2
What am I suppose to find next?

Compare with and .

I am sorry but I do not understand. How am I suppose to compare them?

The values of the function at the critical points are (you needn't verify if these points are local maximum or minimum). The values of at the endpoints of the closed interval are . Comparing those values we obtain the absolute maximum and minimum of that is, and .