# Thread: Absoulute max and min

1. ## 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

2. ## Re: Absoulute max and min

Originally Posted by NeoSonata
f'(x)= 4x^3 -4x = 0
4x(4x^2-1) =0
$4x(x^2-1)=0$

3. ## Re: Absoulute max and min

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

What am I suppose to find next?

4. ## Re: Absoulute max and min

Originally Posted by NeoSonata
so f(0)=3 f(-1)=2 f(1)=2
What am I suppose to find next?
Compare with $f(-2)$ and $f(3)$ .

5. ## Re: Absoulute max and min

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

6. ## Re: Absoulute max and min

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