If the function $f(x)=x^3 + e^{{x}/{2}}$ and $g(x)=f^{-1} (x)$ then find the value of $g'(1)$.

Question

If the function $f(x)=x^3 + e^{\frac{x}{2}}$ and $g(x)=f^{-1} (x)$ then find the value of $g'(1)$.

How to get the inverse of $f(x)$?

Answer 1

We know $g$ is inverse of $f$ so we have $g(f(x))=x$ . Differentiating we have $g'(f(x)).f'(x)=1$ we want $g'(1)$ so in above equation we want $f(x)=1$. Observation suggests that $f(x)=1$ at $x=0$ thus $g'(f(0)).f'(0)=1$ thus $g'(1)\frac{1}{2}=1 $ $g'(1)=2$.

Answer 2

This means $f(0)=1$ or $g(1)=0$. Since $f(x)$ is a bijection $f: R \rightarrow R$ its inverse exists, here it is called $g(x)$. So $$f(g(x))=x \Rightarrow f'(g(x)) g'(x)=1 \Rightarrow g'(1)=\frac{1}{f'(g(1))}= \frac{1}{f'(0)}=2.$$

Note Here the inverse exists but it is not obtainable. Let $y=f(x)$, you can find $g'(y_0)$ only at good points: $(x_0,y_0): (0,1), (1, 1+e^{1/2}), (2, 8+e)..$ for instance $g'(8+e)=1/f'(2)=\frac{2}{24+e}.$ $g'(2)$ exists but it is not obtainable! analytically. This is so becauase $y_0=2$ is not a good point i.e. we do not know the corresponding $x_0$. In general, $$g'(y_0)=\frac{1}{f'(x_0)},$$ where $(x_0, y_0)$ are known good points on the curve $y=f(x)$.

Answer 3

Derivative of inverse function :

$g(f(y))=f^{-1}\circ f (y)=y$ so that chain rule is $$ g'(f(y)) f'(y) =1 $$

Hence $$ f(0)=1,\ g'(1)=g'(f(0))= \frac{1}{f'(0)} =2$$