# Thread: inverse of a function

1. ## inverse of a function

Let $f(x)=\frac{x}{1+|x|}$.
1. Show $f(x)$ is differentiable.
2. Find the inverse of $f(x)$.
3. Show $f^{-1}(x)$ is also differentiable.

I did (1). I have trouble finding the inverse of $f(x)$ so that it is differentiable. Some help please.

2. Hi

$f^{-1}(x) = \frac{x}{1-|x|}$

$f^{-1}\left(f(x)\right) = \frac{x}{1-|\frac{x}{1+|x|}|} = \frac{x}{1-\frac{|x|}{1+|x|}} = x$

3. Originally Posted by running-gag
Hi

$f^{-1}(x) = \frac{x}{1-|x|}$

$f^{-1}\left(f(x)\right) = \frac{x}{1-|\frac{x}{1+|x|}|} = \frac{x}{1-\frac{|x|}{1+|x|}} = x$
But $f^{-1}(x) = \frac{x}{1-|x|}$ is not differentiable at $x=1$ and $x=-1$

4. Originally Posted by dori1123
Let $f(x)=\frac{x}{1+|x|}$.
1. Show $f(x)$ is differentiable.
2. Find the inverse of $f(x)$.
3. Show $f^{-1}(x)$ is also differentiable.

I did (1). I have trouble finding the inverse of $f(x)$ so that it is differentiable. Some help please.
The first thing I'd do is draw a graph of y = f(x).

Note that $f(x) = \frac{x}{1+x} = \frac{-1}{x + 1} + 1$ if $x > 0$ and $f(x) = \frac{x}{1 - x} = \frac{-1}{x - 1} - 1$ if $x \leq 0$.

Clearly the domain of y = f(x) is all real numbers and the range is (-1, 1).

Use the above to inform your thinking on the inverse function and its differentiability. In particular, what's the domain of $y = f^{-1}(x) \, .... ?$

5. The domain for $y = f^{-1}(x)$ is $(-1, 1)$, so $f^{-1}(x)$ is differentiable since $f^{-1}(x)$ is differentiable on $(-1,1)$, right?

6. Originally Posted by dori1123
The domain for $y = f^{-1}(x)$ is $(-1, 1)$, so $f^{-1}(x)$ is differentiable since $f^{-1}(x)$ is differentiable on $(-1,1)$, right?
Correct. I hope you checked for differentiability at the point (0, 0) ....