# Thread: Abs. Max and Min.

1. ## Abs. Max and Min.

I'm having trouble finding the absolute max and min. This is the question:

Find the absolute maximum and absolute minimum values of f(x)=(x^(2/3))*((x^2)-4)

This is what I have so far:

F'(x)=(8/(3(x^1/3)))*((x^2)-1)

Critical #'s I got: x=0(undefined), x=1, x=-1.

Now I don't know if I got the right critical #'s and also I dont know if I have to plug them back into the f(x) or f'(x).

Any help?

2. ## Re: Abs. Max and Min.

Yes, you have the correct derivative and critical numbers. You may want to use the fact that the given function is even to simplify things a bit, i.e., you need only analyze the function for non-negative values of x. So use either the first or second derivative test to determine the nature of the extremum, and look at the limit of the function as x goes to infinity. What do you find?

3. ## Re: Abs. Max and Min.

Sorry I forgot to post that the interval is in between [-2,3]

4. ## Re: Abs. Max and Min.

Well whenever I plug my critical numbers backs into the first derivate i get that when i plug in x=1 then its positive. When I plug in x= -1 its negative. So this means in between the two there is a minimum?

5. ## Re: Abs. Max and Min.

Ah, well then you simply need to evaluate the function at the end-points of the interval, as well as at the critical numbers. The smallest of these values is the absolute minimum and the largest of these values is the absolute maximum. What does the fact that the derivative is not defined at x = 0, but the function is suggest?

6. ## Re: Abs. Max and Min.

That there is a point there but its not differentiable? I don't get what you mean by evaluate. I have the critical numbers and the endpoints. Do I plug them back into f(x) or f'(x)? I'm confused? Lol

7. ## Re: Abs. Max and Min.

This means there is a "cusp" at this point, given the symmetry across the y-axis that an even function has.

When I say evaluate the function at the critical numbers and the end-points, I mean plug these numbers in for x into the original function. These are the only points at which extrema can occur on the given interval.

8. ## Re: Abs. Max and Min.

Okay yes cusp is what I was looking for.

Okay so I plugged in all my critical numbers and endpoints back into the original equation.

This is what i got:
f(1)=-3
f(-1)=-3
f(0)=0
f(-2)=0
f(3)=10.4
So these points show where the max and min occur?
Abs min: f(-1)=f(1)=-3?
Abs max: f(3)=10.4?

9. ## Re: Abs. Max and Min.

Unless you are asked to use a decimal approximation, I would give the absolution maximum as $5\sqrt[3]{9}$.

10. ## Re: Abs. Max and Min.

Ah, it doesn't ask for decimal approximation. Did you get 5sqrt[3](9)? Or is that just more accurate than 10.4?

11. ## Re: Abs. Max and Min.

That's the exact value of f(3):

$f(3)=(3)^{\frac{2}{3}}(3^2-4)=\sqrt[3]{3^2}\cdot5=5\sqrt[3]{9}$