Question

f(x)=x^3, a=8
Find (f^-1)'(a)

Answer 1

well,
f'(x) = 3x^2
f(0)=0

Answer 2

You're supposed to know this: \(\dfrac{d}{dx}f^{-1}(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)}\)

Answer 3

hmm Yes. I know this. I'm just not sure how to apply it xD
I find f^-1(x) first?

Answer 4

no not really

Answer 5

i mean you can if you want, but you don't have to
since \[\dfrac{d}{dx}f^{-1}(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)}\] you know \[\dfrac{d}{dx}f^{-1}(8) = \dfrac{1}{f'\left(f^{-1}(8)\right)}\]

Answer 6

you can find \(f^{-1}(8)\) easily by solving \[x^3=8\] fror \(x\)

Answer 7

x=2

Answer 8

So then we have
\[\frac{ 1 }{ f'(2) }=\frac{ 1 }{ 12 }\]

Answer 9

yay good job

Answer 10

Yaay. Thanks @satellite73 :)