locate absolute extreme of h(t)=(t)/(t-2) on closed interval [3,5]

Question

tkhunny · Answer

Have you considered the 1st derivative?  That will help you on your way, but the derivative doesn't know that chopped it off at 3 and 5.

anonymous · Answer

can yu walk me through the steps?

tkhunny · Answer

1st Derivative is all yours.  Let's see what you get.

anonymous · Answer

$$\frac{ 2 }{ (t-2) ^2}$$

anonymous · Answer

from what i know of i tink i set that to 0 and find t right?

tkhunny · Answer

First, you get the derivative right.

$h(t) = \dfrac{t}{t-2}$

$h'(t) = \dfrac{(t-2)(1) - t(1)}{(t-2)^{2}} = \dfrac{t - 2 - t}{(t-2)^{2}} = \dfrac{-2}{(t-2)^{2}}$

You should see the value of showing your work.  Just pulling things out of the sky is not very useful.

anonymous · Answer

oh i forgot the negative

anonymous · Answer

and now i set it to 0?