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

$\displaystyle 4x(x^2-1)=0$

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 $\displaystyle f(-2)$ and $\displaystyle f(3)$ .

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 $\displaystyle f(0)=3,f(1)=f(-1)=2$ (you needn't verify if these points are local maximum or minimum). The values of $\displaystyle f$ at the endpoints of the closed interval are $\displaystyle f(-2)=11,f(3)=66$ . Comparing those values we obtain the absolute maximum and minimum of $\displaystyle f$ that is, $\displaystyle f_{\max}(3)=66$ and $\displaystyle f_{\min}(1)=f_{\min}(-1)=2$ .