For each of the piecewise-defined functions, determine whether or not the function is one-to-one, and if it is, determine its inverse function.f(x)={(x , when x < 0),(2x, when x >=0):}

Step 1
When x is less than 0 then also y is less than zero
if ${\displaystyle x_1\ne x_{2}f\text{or}{x}_1,x_2<0}$
Then, ${\displaystyle y_1\ne y_2}$
if ${\displaystyle x\ge 0}$ , then y is also greater than 0.
if ${\displaystyle x_3\ne x_{4}f\text{or}{x}_3,x_4<0}$
Then ${\displaystyle 2x_3\ne 2x_4}$
So, ${\displaystyle y_3\ne y_4}$
This means it is a one-to-one function
Step 2
To find the inverse functions we switch x and y and then solve for y
For, ${\displaystyle x<0}$
$y=x$
Or, $x=y$
For ${\displaystyle x\ge 0}$
$y=2x$
Or, $x=2y$
Or, ${\displaystyle y=\frac{x}{2}}$
Result:
${\displaystyle f^{-1}\left(x\right)=\{\begin{array}{cc}x& x<0\\ \frac{x}{2}& x\ge 0\end{array}}$