Write the function $f(x) = 0, \text{if $x \leq 0$}; x, \text{if $x > 0$}$ as a single formula using the absolute value sign. I found this exercise in Demidovich's "Problems in Mathematical Analysis", and found it quite interesting.

Write the function
  $$
f(x) =
\begin{cases}
0, & \text{if $x \leq 0$} \\
x, & \text{if $x > 0$}
\end{cases}
$$
  as a single formula using the absolute value sign.

I've posted my solution below.
 A: Answer
$$f(x) = \frac{x + \lvert x \rvert}{2}$$

Explanation
When $x \leq 0$, $x + \lvert x\rvert = 0$, rendering the $\text{RHS} = 0$.
When $x > 0$, $x + \lvert x \rvert = 2x$. So, it must be divided by $2$ for $\text{RHS} = x$.
A: It seems natural to generalize this problem to:

Write the function
  $$
f_{ab}(x) =
\begin{cases}
ax, & \text{if $x \leq 0$} \\
bx, & \text{if $x > 0$}
\end{cases}
$$
  as a single formula using the absolute value sign.

Having done so:


*

*Notice that the slope of $f_{ab}$ jumps by $(b-a)$ at $0$; $f_{ab}$ is smooth elsewhere. The slope of the absolute value function changes by $2$ at $0$, suggesting that we consider the function
$$
g_{ab}(x)=f_{ab}(x)-\frac{b-a}{2}|x|
$$
which will "smooth out" $f$ at $0$.

*Having noticed this, it's not too hard to check by case analysis that in fact
$$
g_{ab}(x)=\frac{b+a}{2}x
$$
and so
$$
f_{ab}(x)=\frac{b+a}{2}x + \frac{b-a}{2}|x| 
$$
When $a=0$ and $b=1$ as in the original problem, this reduces to the solution in the other answer.
