# Critical Values

1. Nov 11, 2005

### scorpa

Hello Everyone,

I'm doing a question on critical values and absolute max and mins and for some stupid reason I can't seem to get one of the questions that was assigned.

Find the absolute max and/or min of the (cube root t )(8-t) in the closed interval [0,8]

First to solve this found the first derivative so that I could find the critical numbers of the equation, which is where the first derivative equals zero.
So when I found the first derivative I got ((1/3)(8-t)-t^(2-3))/(cube root t)

Now I know that one of these critical values is obviously going to be at t=0. But I'm stumped when it comes to finding the critical number from the numerator. I know this is a really stupid question, but I just can't seem to figure it out. Thanks in advance for any advice you can give.

2. Nov 11, 2005

### Fermat

Your derivative seems to be a bit off.

$$\mbox{If } f(t) = t^\frac{1}{3}\cdtot (8-t)$$

$$f'(t) = \frac{1}{3}\cdot t^\frac{-2}{3}\cdot (8-t) - t^\frac{1}{3} = 0$$
$$f'(t) = \frac{(8-t)}{3t^\frac{2}{3}} - \frac{3t}{3t^\frac{2}{3}} = 0$$

3. Nov 11, 2005

### scorpa

Oh geez, I can't believe myself. I wrote the wrong exponent when I was doing the first derivative, I can be such an idiot sometimes. Thanks a lot, the question is a lot easier now...haha.