Question

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

Piecewise-Defined Functions

For each of the piecewise-defined functions in determine whether or not the function is one-to-one, and if it is, determine its inverse function.
$$\displaystyle{f{{\left({x}\right)}}}={\left\lbrace\begin{array}{cc} {x}^{{2}}&{w}{h}{e}{n} \ {x}{<}{0}\\{x}&{w}{h}{e}{n} \ {x}\ge{0}\end{array}\right.}$$

2020-11-09

Step 1
Let $$\displaystyle{x}{<}{0}$$
Then we get,
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}$$
Let us consider two numbers $$\displaystyle{x}{<}{0}{\quad\text{and}\quad}{y}{<}{0}$$.
Also let us assume,
$$\displaystyle{x}^{{2}}={y}^{{2}}$$
$$\displaystyle{x}^{{2}}-{y}^{{2}}={0}$$
(x + y)(x - y) = 0
(x + y) = 0, (x - y) = 0
x = -y, x = y
Therefore for two elements x and -x we have $$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}$$, as well as $$\displaystyle{f{{\left(-{x}\right)}}}={\left(-{x}\right)}^{{2}}={x}^{{2}}={f{{\left({x}\right)}}}$$.
Hence f is not one-to-one for $$\displaystyle{x}{<}{0}$$.
Step 2
Let $$\displaystyle{x}\ge{0}$$
Then we get,
f(x) = x
Let us assume for two elements x and y both greater than or equal to zero, their images are same.
So we get,
f(x) = f(y)
x = y
So for $$\displaystyle{x}\ge{0},{y}\ge{0}$$, f(x) = f(y) implies x = y.
Hence f is one-to-one function.