Find $a$ such that the piecewise function $f $ is injective and $a$ such that $f$ is surjective Let $f(x)= \begin{cases} 
      ax+2, & x\leq 1 \\
       \
      x+2a, & x\gt 1 
   \end{cases}
$ where $a\in\mathbb{R}$
Find $a$ such that:
$f$ is injective
$f$ is surjective
EDIT: I need $a$ so $f$ is injective first and then surjective, not both at the same time.
I usually study injectivity and surjectivity with the first derivative, but in this case I do not think i could use that because i would have to assume the function is differentiable at $x=1$. So I tried taking 3 cases for injectivity:
$f(x) = f(y)$ where $x\leq1$ and $y\leq1$
$f(x) = f(y)$ where $x\gt1$ and $y\gt1$
$f(x) = f(y)$ where $x\leq1$ and $y\gt1$
The first 2 cases were easy to solve but i am stuck at the third case as I have too many unknowns and I do not know what to do.
For surjectivity, I tried taking limits at $+\infty$, $1$ and $-\infty$ to figure out the Image of $f$, but aside from limit to $+\infty$ where i get $+\infty$, I still have $a$ in the result for the other limits.
How should i solve this problem ?
Thanks in advance !
 A: For injectivity and surjectivity, I advise finding the range of $f$ when restricted to $(-\infty, 1]$ and when $f$ is restricted to $(1, \infty)$.
The union of these two ranges produces the range of $f$. So, $f$ is surjective if and only if the union of these ranges is $\Bbb{R}$.
If these ranges intersect, then the function is not injective (since there is some function value attained both in $(-\infty, 1]$ and $(1, \infty)$, contradicting injectivity). Moreover, $f$ is injective on both subdomains and these ranges fail to intersect if and only if $f$ is injective over $\Bbb{R}$.
So, let's consider these two ranges. Let $R_1 = f(-\infty, 1]$ and $R_2 = f(1, \infty)$. Considering $R_1$ first, we need to consider certain possibilities for $a$. If $a = 0$, then $f(x) = 2$ for all $x \le 1$, so $R_1 = \{2\}$. If $a > 0$, then $f$ is increasing on $(-\infty, 1]$, and tends to $-\infty$ as $x \to -\infty$, so $R_1 = (-\infty, a + 2]$. Finally, if $a < 0$, then $f$ is decreasing, tending to $\infty$ as $x \to -\infty$, giving us $R_1 = [a + 2, \infty)$. In summary,
$$R_1 = \begin{cases} (-\infty, a + 2] & \text{if } a > 0 \\ \{2\} & \text{if } a = 0 \\ [a + 2, \infty) & \text{if } a < 0. \end{cases}$$
On the other hand, $f$ is always increasing on $(1, \infty)$, with $f(x) \to \infty$ as $x \to \infty$, so
$$R_2 = (1 + 2a, \infty).$$
Where is $f$ surjective? Certainly, if $a \le 0$, then both $R_1$ and $R_2$ are intervals with a lower bound, thus they cannot cover all of $\Bbb{R}$. So, we are considering $a > 0$ from the get-go. In order for the two intervals $(-\infty, a + 2]$ and $[1 + 2a, \infty)$ to cover $\Bbb{R}$, we need $a + 2 \ge 1 + 2a \iff a \le 1$. So, $f$ is surjective if and only if $0 < a \le 1$.
Now, $f$ is always injective on $(1, \infty)$, and $f$ is injective on $(-\infty, 1]$ if and only if $a \neq 0$. Note that if $a < 0$, then both intervals tend are unbounded above, and thus non-trivially intersect. This implies $f$ is not injective if $a \le 0$.
If $a > 0$, then we need the two intervals not to intersect, which occurs precisely when $a + 2 \le 1 + 2a \iff a \ge 1$.
Thus, in conclusion, $f$ is injective when $a \in [1, \infty)$ and is surjective when $a \in (0, 1]$. It is bijective precisely when $a = 1$.
