I have encountered a problem which I'm unsure if I've approached correctly. Any input as to whether or not my approach is correct, is greatly appreciated. Note that English is not my first language, so I apologize in advance for any typos.
Problem: Given the function:
where $k$ is a real number. For what values of $k$ does the function $f$ have an inverse function?
Attempted solution:
First it should be pointed out that this function is continuous no matter which value we choose for $k$.
I then begin by finding the derivative of each part of the piecewise function:
For the function to have an inverse, it must be injective. I therefore begin by investigating where each part of the function will have a potential maximum or minimum. That is, where $f'(x) = 0$.
For the first part of the piecewise function, we then get:
$-2x + 2 + k = 0$
Solving for $x$ yields:
$x = \frac{k}{2} + 1$
For the second part of the piecewise function, we get:
$2x + 2 - k = 0$
Solving for $x$ yields:
$x=\frac{k}{2} - 1$
We see from that this that the maximum and minimum values we obtain will depend on which value we choose for $k$.
Here is where I am a little bit unsure about how to proceed. I tried to reason as follows: The two parts of the function will meet at $x=k$. So let us find out for which values of $k$ the maxima or minima for one part of the function will coincide with the other part of the fuction. We then obtain, for the first solution:
$k = \frac{k}{2} + 1$
Solving for $k$ yields:
$k = 2$
For the second solution, we obtain:
$k = \frac{k}{2} - 1$
Here we obtain:
$k = -2$
Lastly, I investigate for each of the intervals, $k < -2, -2\leq k \leq 2$ and $k > 2$ where we obtain similar signs for the derivative for each of parts of the piecewise function. I then discovered that this occurs for $-2 \leq k \leq 2$. So by choosing $k$ in this interval, we will have an inverse function as the function will then be injective.
I am, however, unsure if my reasonsing here is correct and if my solution is rigorous enough. Particularly, when I set $x=k$ for where we have a minima or maxima. If someone can explain to me if my solution and reasoning is sound, then I would greatly appreciate it! Thanks in advance!
PS: I did not know how to write the piecewise functions in Latex, so I hope it is okay that I uploaded these as images.
Yeah your reasoning is correct to me. When $|k|>2$, either left part or right part of $f$ is not injective. When $|k|\leq2$, you can show $f'(x)>0$ for any $x\neq k$, then by continuity you know $f$ is injective.