Can't find the inverse!

Question

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.

Answer 1

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.

Answer 2

Use the formula $(f^{-1})^{'}(x) = \frac 1 {f^{'}(f^{-1}(x))}$ then use the fact that $f^{-1}(1)=0$

Answer 3

Use the chain rule to differentiate both sides of either $$f(g(x))=x$$or $$g(f(x))=x$$whichever seems easiest to you.

Answer 4

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:$$g'(1)f'(0)=1.$$

Answer 5

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.

Answer 6

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.

Answer 7

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.