For which values of $ \alpha \in R$ the function $f(x)= x+ \alpha \cdot |x|$ is invertible I have to say for which values of $ \alpha \in R$ the function $f(x)= x+ \alpha \cdot|x|$ is invertible.
$$f(x)=\begin{cases} (1+ \alpha)x &  x \ge 0\\
(1- \alpha)x &  x < 0 \end{cases}$$
$$D=R$$
If $x_1, x_2 \in [0, + \infty), f(x_1)=f(x_2) \Rightarrow x_1=x_2 $
If $x_1, x_2 \in (- \infty, 0], f(x_1)=f(x_2) \Rightarrow x_1=x_2 $
If $x_1 \in [0, + \infty), x_2 \in (- \infty, 0], f(x_1)=f(x_2) \Rightarrow x_1= \frac{1-\alpha}{1+ \alpha}x_2 $
If $x_1 \in (- \infty, 0], x_2 \in [0, + \infty), f(x_1)=f(x_2) \Rightarrow x_1= \frac{1+\alpha}{1- \alpha}x_2 $
So the function is always injective  in $(- \infty, 0]$ and $[0, + \infty)$ but it is injective in all its domain only of $\alpha =0$.
That's my result, while in my book it says that the function is injective $ \forall \alpha \in R $ so that $| \alpha |<1$
Can someone show me where I made mistakes?
 A: $\frac{1-\alpha}{1+\alpha}$ should be negative (since $x_1$ and $x_2$ are of opposite sign) this leads to $|\alpha|\ge 1$.
So injectivity when this is not realized (i.e. $|\alpha|<1$). Same for the other one.
Maybe you can go directly for $f'$ sign and discuss when $f$ is strictly monotonical instead.
A: In your third case, where $x_1 \in [0, + \infty), x_2 \in (- \infty, 0]$, i.e. $x_1$ and $_2$ have different signs, you deduce that $x_1= \frac{1-\alpha}{1+ \alpha}x_2$. But for some values of $\alpha$ this leads to a contradiction, because $\frac{1-\alpha}{1+ \alpha}$ is positive.
So you should look for values of $\alpha$ where it's impossible for both "different signs" and $x_1= \frac{1-\alpha}{1+ \alpha}x_2$ to be true simultaneously — $f$ will be invertible then.
A: You assume that there exist $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$ and $x_1$ and $x_2$ are of opposite sign. From there you conclude that $x_1 = \frac{1-\alpha}{1+\alpha} x_2$ or $x_2 = \frac{1-\alpha}{1+\alpha} x_1$.
Now, you reason that $x_1$ and $x_2$ should be equal for the function to be injective and therefore $\alpha = 0$. But this is a wrong thing to do since you already assumed that $x_1$ and $x_2$ are of opposite sign (I should be more pedantic and treat $0$ separately, but I won't).
Rather, you want to show that if $x_1$ and $x_2$ are of opposite sign, $f(x_1) = f(x_2)$ is impossible for certain $\alpha$'s. This corresponds to the characterization of injectivity that says:
$$x_1\neq x_2 \implies f(x_1)\neq f(x_2),\quad \forall x_1,x_2\in\mathbb R.$$
So, you assume $x_1\neq x_2$. If $x_1$ and $x_2$ are of the same sign, it follows that $f(x_1) \neq f(x_2)$ since you know $f$ is injective on $(-\infty , 0]$ and $[0,\infty)$ (when $|\alpha| \neq 1$). What remains is the case when $x_1$ and $x_2$ are of opposite signs, without loss of generality, $x_1 > 0$. Then, $f(x_1) = f(x_2)$ if and only if $(1+\alpha) x_1 = (1-\alpha)x_2$.
We arrived at the important part. If $1+\alpha$ and $1-\alpha$ are of opposite signs, we are done, injectivity fails. If they are of the same sign, however, we arrive at contradiction with the assumption that $x_1$ and $x_2$ are of opposite signs and $f(x_1) = f(x_2)$ is impossible. This would complete the proof that $x_1 \neq x_2$ implies $f(x_1)\neq f(x_2).$
It turns out that $f$ is injective if and only if $1+\alpha$ and $1-\alpha$ are of the same sign, and none is $0$. This is equivalent to $|\alpha|<1$. This is actually expected result, since you have two lines patched together. The result is a graph of injective function unless the lines have opposite signed slopes.
Note that I skipped the edge cases of $x = 0$ or $|\alpha| = 1$. For a complete proof, you shouldn't do that. Hopefully, this clarified things for you.
