Let f (x) = x^3+x and let g=f^(-1)
(you may take it as given that an inverse exists). Find g'(0) and g'(10).

Question

Let f (x) = x^3+x and let g=f^(-1)
(you may take it as given that an inverse exists). Find g'(0) and g'(10).

anonymous · Answer

that was wrong
$$\frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'(f^{-1}(x))}$$

anonymous · Answer

you want the derivative of the inverse ( here called $g$ ) at $x=0$so we need two things, $f^{-1}(0)$ and $f'(x)$

anonymous · Answer

$$f^{-1}(0)$$ solve 
$$f(x)=0$$ for  $x$ via 
$$x^3+x=0\implies x=0$$ so 
$$f^{-1}(0)=0$$

anonymous · Answer

$$f'(x)=3x^2+1$$ and therefore 
$$f'(f^{-1}(0))=f'(0)=1$$ and so your answer is $\frac{1}{1}=1$

anonymous · Answer

Thank you very much! Does that mean:
To find g'(10):
$$solve, x^3+x=10$$
$$\therefore x=2$$
$$f^{-1}(2)=13$$
$$\therefore g'(10)=1/13$$

anonymous · Answer

x=y^3+y ,(x=0,y=0), (x=10,y=2) so  1=3y'y^2+y'  then  y'=1/(3y^2+1)= 1/1 ,  1/13

anonymous · Answer

yes, you got it!