Help would be much appreciated ^.^
-Find the absolute and local maximum and minimum values of f(t) = 2/t + 10 , 0 < t ≤ 4

- anonymous

- chestercat

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- myininaya

Well we know we need to find f'(t)
is f(t)=2/t+10 or f(t)=2/(t+10)?

- anonymous

the 1st one :)

- myininaya

just so i can know to check your derivative

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## More answers

- myininaya

ok cool

- myininaya

\[f(t)=\frac{2}{t}+10 \\ f(t)=2t^{-1}+10 \\\]
do you know how to find f'?

- anonymous

Yes! It's \[f'(t)= \frac{ -2 }{ t^2 }\] am I right?

- myininaya

yes and f' is not zero ever
but f' dne at t=0 but we f(0) also doesn't exist
at t=0 we actually have a vertical asymptote
|dw:1434748364786:dw|
so we don't have to worry about critical numbers but this just shows we also don't have any max of any type
but w do have a min and it is an absolute min (you know since local min can't occur at endpoints)

- myininaya

so you just need to consider the endpoints of your interval given

- myininaya

tell me if you can find the absolute min?

- anonymous

Ohhh I understand.
How can I find the absolute max. and min. ?

- myininaya

there is a vertical aysmptote in the pic and as x approaches it from the right the y values keep getting bigger and bigger there is no max
you only need to find the absolute min
and I gave you a big hint for that it occurs at the endpoint for this question

- myininaya

|dw:1434748634077:dw|

- anonymous

Oh yeah! So I plug in 4 into the original function and that will be my abs. min. ?

- myininaya

yes

- anonymous

So that means it is 10.5 ?

- myininaya

yes

- myininaya

no calculus was actually needed on this problem

- anonymous

If I did not known what the graph looked like would I still be able to find it?

- myininaya

Yes.

- myininaya

I said no calculus was needed because g(x)=1/x is a pretty basic graph learned in algebra. f(x)=2/x+10 is just a translation of that graph.
Anyways f(x)=2/x+10 , 0

- anonymous

@myininaya thank you so much!!!

- myininaya

Means there is no abs max at all.*
Since there were no critical numbers then there are no relative max/min

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