Can't find the inverse! If $$f(x) = x^2 + x + e^x\\g(x) = f^{-1}(x) $$
then use the chain rule to find $g'(1)$.
When I tried to find the inverse of $f(x)$ I couldn't as it is not a one-to-one function so I could not solve it.
I'd appreciate your help.
 A: You need not find the inverse. The computation simply requires application of the chain rule.
Observe that
$$g(x)=f^{-1}(x)$$
$$\Rightarrow f\{g(x)\}=x$$
Now, differentiating both sides with respect to $x$, using chain rule, we get
$$\frac{d}{dx}f\{g(x)\}=\frac{d}{dx}x$$
$$\Rightarrow \frac{d}{dg(x)}f\{g(x)\}\cdot \frac{dg(x)}{dx}=1$$
$$\Rightarrow f'\{g(x)\}\cdot g'(x)=1$$
$$\Rightarrow g'(x)=\frac{1}{f'\{g(x)\}}$$
Hence $$g'(1)=\frac{1}{f'\{g(1)\}}$$
Hope you can finish this on your own.
A: Since $g\bigl(f(x)\bigr)=x$, the chain rule implies that$$g'\bigl(f(x)\bigr)f'(x)=1.$$In particular, put $x=0$ and you will get:$$g'(1)f'(0)=1.$$
A: Use the formula $(f^{-1})^{'}(x) = \frac 1 {f^{'}(f^{-1}(x))}$ then use the fact that $f^{-1}(1)=0$
A: Use the chain rule to differentiate both sides of either $$f(g(x))=x$$or $$g(f(x))=x$$whichever seems easiest to you.
A: A nice little trick:
Notice the following
$$\frac{d}{dx} f(f^{-1}(x)) = \frac{d}{dx}(f^{-1}(x)) \cdot  f'(f^{-1} (x))$$
where the equality comes from the chain rule. On the left hand side, however, we know that 
$$f(f^{-1}(x)) = x$$
because they are inverse functions. Therefore, 
$$\frac{d}{dx} f(f^{-1}(x)) = \frac{d}{dx} x = 1$$
So putting this into the first equation, we have that 
$$1 = \frac{d}{dx}(f^{-1}(x)) \cdot  f'(f^{-1} (x))$$
$$\implies \frac{d}{dx}(f^{-1}(x)) = \frac{1}{f'(f^{-1} (x))}$$
Since $g(x) = f^{-1}(x)$ then we have
$$g'(x) = \frac{1}{f'(g(x))}$$
$$\implies g'(1) = \frac{1}{f'(g(1))}$$
If we do a bit of a guess, we see that $f(0) = 1$. I'll leave it from here for you to do. 
A: This function hasn't an inverse function that you can describe using elementary functions. If you have a given value, $ y_0 $ and you want to know for which $ x_0 $ is true the equation $ y_0 = f(x_0) $ you may use the Newton's method: Newton's method in Wikipedia.
A: It is true the function $f$ is not one-to-one for all $x$. Instead one must restrict its domain in order for the inverse to exist at the point of interest of $x = 1$. 
As the function $f$ has a single global minimum at the point 
$$x^* = -\frac{1}{2} - \text{W}_0 \left (\frac{1}{2 \sqrt{e}} \right ) = -0.738\,835\ldots,$$
where $\text{W}_0(x)$ is the principal branch of the Lambert W function, by restricting the domain of the function $f$ to $x \geqslant x^*$ an inverse $g (x) = f^{-1}(x)$ containing the point $x = 1$ exists though an explicit expression for it cannot be found.
It may however still be possible to find individual values for $f^{-1} (x)$ at particular values for $x$. In our case we note that $f^{-1} (1) = 0$. This allows for the value of the derivative at this point to be found using
$$(f^{-1})'(1) = \frac{1}{f'(f^{-1} (1))} = \frac{1}{f'(0)},$$
since we know that
$$(f^{-1})' (x) = \frac{1}{f'(f^{-1} (x))},$$
and the value $f'(0)$ can be readily found. 
