Question

Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results.

h(t) = (t)/((t − 3)^2) , [4, 7]

Absolute Max (?,?)
Absolute Min (?,?)

Answer 1

First step is to differentiate h.

Answer 2

using the quotient rule? which would be \[\frac{ -3 }{ (t-3) ^2}\]

Answer 3

hmm.... this is the quotient rule applied to your h function:
\[h'(t)=\frac{(t)'(t-3)^2-t[(t-3)^2]'}{[(t-3)^2]^2}\]

Answer 4

I want to save you pain. Do not multiply anything out at this moment. Just differentiate those factors with the ' on them.

Answer 5

\[-\frac{t+3 }{(t-3)^3 }\]

Answer 6

(t)'=1
[(t-3)^2]'=2(t-3)

Answer 7

on you canceled a factor
let me recheck it

Answer 8

yeah still missing something

Answer 9

no you did fine lol

Answer 10

I wasn't expecting you to simplify
but okay lol

Answer 11

Ok and we have critical numbers t=-3
which doesn't happen in the interval [4,7]

Answer 12

so the only thing you need to look at are endpoints

Answer 13

loso i have to plug in the endpoints and that will give me the values ?

Answer 14

\[h(4)=? \\ h(7)=? \]

Answer 15

yep

Answer 16

If we had any critical numbers to occur in between 4 and 7
we would plug in those as well but we didn't

Answer 17

oh alrighty thank you !

Answer 18

You got it from here?

Answer 19

i think so for h(4) i got -7 for h(7) i got um kinda weird -10/64?

Answer 20

\[h(4)=\frac{4}{(4-3)^2} \\ h(7)=\frac{7}{(7-3)^2}\]

Answer 21

oh i plugged it in the derivative thats why

Answer 22

The derivative was just to give us any critical numbers between 4 and 7
The function that is actually going to tell where (4,h(4)) and (7,h(7)) are going to be is h

Answer 23

i got for h(4)=4 and for h(7)=7/16

Answer 24

i got the absolute maximum right (4,4)
but i got the absolute minimum wrong (7, 7/16)

Answer 25

sounds good to me