# 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):}

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.

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Step 1
When x is less than 0 then also y is less than zero
if ${x}_{1}\ne {x}_{2}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{x}_{1},{x}_{2}<0$
Then, ${y}_{1}\ne {y}_{2}$
if $x\ge 0$ , then y is also greater than 0.
if ${x}_{3}\ne {x}_{4}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{x}_{3},{x}_{4}<0$
Then $2{x}_{3}\ne 2{x}_{4}$
So, ${y}_{3}\ne {y}_{4}$
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, $x<0$
$y=x$
Or, $x=y$
For $x\ge 0$
$y=2x$
Or, $x=2y$
Or, $y=\frac{x}{2}$
Result:
${f}^{-1}\left(x\right)=\left\{\begin{array}{cc}x& x<0\\ \frac{x}{2}& x\ge 0\end{array}$