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

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

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

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

you just want to sent the equation equal to 0?

oops set

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

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

correct

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

oh! I completely overlooked the inverse over there, sorry

yeah those lovely inverses

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

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

ok

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

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

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

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

Thanks for your help slaaibak.
Have a great night.

No problem. Thanx, you too.