Question

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.
\(\displaystyle{f{{\left({x}\right)}}}={\left\lbrace\begin{array}{cc} {x}&{w}{h}{e}{n}{x}{<}{0}\\{2}{x}&{w}{h}{e}{n}{x}\ge{0}\end{array}\right.}\)</span>

Answer 1

Step 1
When x is less than 0 then also y is less than zero
if \(\displaystyle{x}_{{1}}\ne{x}_{{2}}{f}{\quad\text{or}\quad}{x}_{{1}},{x}_{{2}}{<}{0}\)</span>
Then, \(\displaystyle{y}_{{1}}\ne{y}_{{2}}\)
if \(\displaystyle{x}\ge{0}\) , then y is also greater than 0.
if \(\displaystyle{x}_{{3}}\ne{x}_{{4}}{f}{\quad\text{or}\quad}{x}_{{3}},{x}_{{4}}{<}{0}\)</span>
Then \(\displaystyle{2}{x}_{{3}}\ne{2}{x}_{{4}}\)
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, \(\displaystyle{x}{<}{0}\)</span>
y=x
Or, x=y
For \(\displaystyle{x}\ge{0}\)
y=2x
Or, x=2y
Or, \(\displaystyle{y}=\frac{{x}}{{2}}\)
Result:
\(\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}={\left\lbrace\begin{array}{cc} {x}&{x}{<}{0}\\\frac{{x}}{{2}}&{x}\ge{0}\end{array}\right.}\)</span>