Question

Find the absolute max and min of f(x)=t(4-t)^(1/2) on [-1,3]
I can't find the max I took the derivative of t(4-t)^(1/2) and tried to factor it

So far t(1/2(4-t)^(-1/2)(-1)+(4-t)^(1/2)

can't figure out how to get anything out of it

Answer 1

You need to write as one fraction

Answer 2

Try rewriting with positive exponents first

Answer 3

I did that then do I just rationalize?

Answer 4

You don't need to rationalize anything

Answer 5

i just can't see how to factor it out hm...

Answer 6

\[y'=\frac{-t}{2 \sqrt{4-t} }+\sqrt{4-t}=\frac{-t+2\sqrt{4-t} \cdot \sqrt{4-t}}{2 \sqrt{4-t}}\]
Is that what you got?

Answer 7

And of course we can make that a little prettier.

Answer 8

because thats the only way I can see of how to get a value from it, yes that is what I got! :D

Answer 9

I thnik i see it now then jsut solve for the polynomial that its going to yield on top?

Answer 10

\[y'=\frac{-t+2(4-t)}{2 \sqrt{4-t}}=\frac{-t+8-t}{2 \sqrt{4-t}}=\frac{-2t +8}{2 \sqrt{4-t}}\]

Answer 11

hmm thats 4 thought its out of the range

Answer 12

then what would the max be hmmm

Answer 13

4 is an element of the domain of f

Answer 14

that would mean t would have to = 4 but thats not in [-1,3], or am I missing something?

Answer 15

Critical numbers are elements of the domain of f that satisfy
f'=0
or f' DNE

Answer 16

but what if they are outside the given interval?

Answer 17

regardless it wouldnt be a max becasue it would make it 0 hm...