Inverse function

[Imum Coeli]

I need some help solving this problem.

Question:
Let f(x) = sin(pi/2*x)+x , (0<=x<=1) and let g = f^-1 (It may be taken as a given that an inverse exists). Find g'(0) and g'(2).

Notes:
I thought that I could use g'(x) = 1/(f'(g(x))) because I can find f'(x) but I still need g(x).

My method is:

let y = sin(pi/2*x)+x
then I believe that
arcsin(y) = pi/2*x + arcsin(x)
But I still can't figure out how to isolate x to find the inverse.

I'm not sure if I'm even approaching this the right way at all. In other words I am totally lost.
Thanks for any help.

 

[pka]

Question:
Let f(x) = sin(pi/2*x)+x , (0<=x<=1) and let g = f^-1 (It may be taken as a given that an inverse exists).
Find g'(0) and g'(2).

The key to this question is the the realization that \(\displaystyle f(0)=0~\&~f\left( 1 \right) = 2\).

Therefore, \(\displaystyle g'(0)=\dfrac{1}{f'(0)}~\&~g'(2)=\dfrac{1}{f'\left( {f(1)} \right)}\)

 

[Imum Coeli]

Oh thanks. Very clever.
Thank you so much. (Doesn't sin(pi/2*x)+x = 2 when x=1?)

 

[pka]

(Doesn't sin(pi/2*x)+x = 2 when x=1?)

Correct. Edited