- anonymous

If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)

- katieb

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- anonymous

\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\]
find \[(f ^{-1})'(0)\]

- anonymous

\[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\]
\[f(x) = (1+x^3)^{1/2}\]
\[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{-1/2}} (3x ^{2})\]

- slaaibak

f'(x) = sqrt(1+x^3)
so f'(0= 1

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## More answers

- anonymous

is that setting up:
putting the 0 from (f^-1)'
\[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]

- slaaibak

no.
so f'(0)= 1 meaning sqrt(1+0^3) = 1

- anonymous

you just want to sent the equation equal to 0?

- anonymous

oops set

- slaaibak

bro, you asked me to find f'(0) and I found f'(0), what are you on about?

- slaaibak

you know evaluating a function at zero and setting it equal to zero are two different things?

- anonymous

correct

- slaaibak

so you gave f(x). I found f'(x)
and then I evaluated it at zero (f'(0))
That's the question, isn't it?

- anonymous

Yes we evaluated f'(x) at zero
and we ant to put that into \[(f ^{-1})'(0) = \frac{ 1 }{ f'(f ^{-1}(0) }\]
Correct? Or am I too tired and my brain is done for the day.

- anonymous

\[= \frac{ 1 }{ f'(1) }\]

- slaaibak

oh! I completely overlooked the inverse over there, sorry

- anonymous

yeah those lovely inverses

- anonymous

do we need to back track some for the steps done for doing f'(0) or are we good?

- slaaibak

so yeah, you solve 0 = f(x)
then x=3
do you agree? Then f^-1(0) = 3 ? or alternatively, f(3) = 0

- anonymous

ok

- slaaibak

then f'(3) = sqrt(1+3^3) = sqrt(28)

- anonymous

dang I didn't need to figure that ugly f'(x)?

- slaaibak

you did. Well, I did.
\[f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}\]

- slaaibak

it's kinda obvious if you think about it. you should get the same answer back, since you are integrating and then differentiating again. BUT there is a chain rule adjustment that has to be made if there are more complex bounds

- anonymous

I wish these derivative inverses made more sense to me.
I am pretty confident in my inverse solving. But these inverse derivatives are killing me.

- slaaibak

haha, just remember the methods/formulas, do it step by step and you'll be fine

- anonymous

Thanks for your help slaaibak.
Have a great night.

- slaaibak

No problem. Thanx, you too.

- anonymous

ok can some one work this problem for me from 1st step to last step. I see you plug x in for t. and 3 in for x but what about the (f^-1)(0) what does that fit in
thanks

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