Question

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

Answer 1

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.

Answer 2

can yu walk me through the steps?

Answer 3

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

Answer 4

\[\frac{ 2 }{ (t-2) ^2}\]

Answer 5

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

Answer 6

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.

Answer 7

oh i forgot the negative

Answer 8

and now i set it to 0?