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
ANSWERED
asked 2020-11-08

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

Answers (1)

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.

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