derivative of an inverse function that is a cubic, HELP !

[sinclairharry]

I've tried very long trying to figure out this question and I just can't !

Find (f^-1)'(9) if f(x) = x^3 + x - 1

I thought the answer could be

x = y^3 + y - 1

dx/dy = 3y^2 + 1

dy/dx = 1 / (3y^2 + 1)

swap x and y again so (f^-1)' = 1 / (3y^2 + 1)

then

(f^-1)'(9) = 1/ (3*(9)^2 + 1) = 1 / 244

but that seems to be incorrect
I seem to be a bit lost any advice ?

 

[pka]

I've tried very long trying to figure out this question and I just can't !
Find (f^-1)'(9) if f(x) = x^3 + x -1

Stop right there. First you need to answer this. If \(\displaystyle f(a)=9\) then what is \(\displaystyle a=~?\)

Then \(\displaystyle [f^{-1}]^{\prime}(9)=\dfrac{1}{f'(a)}\)

 

[sinclairharry]

Stop right there. First you need to answer this. If \(\displaystyle f(a)=9\) then what is \(\displaystyle a=~?\)

Then \(\displaystyle [f^{-1}]^{\prime}(9)=\dfrac{1}{f'(a)}\)

Oh ok !

So if f(a) = 9 then a = 2

so 1 / 3 (2)^2 + 1 = 1 / 13

which is the answer !

thank you so much !