Chapter 1, Problem 58CRE

To determine

Whether the below function is an increasing function and find out if the inverse of the function exists. If the inverse exists, find out the inverse of the function.

$f\left(x\right)=\left\{(x\text{ - 1) }whenx<0andlnxwhenx\ge 0\right\}$

To determine

To find if the inverse of the function exists:

For the interval $(\infty ,0)$, $f\left(x\right)=x\text{ - 1}$

Let us assume that if $f\left(-a\right)=f\left(-b\right)$

Then, $a\text{ - 1 }=b\text{ - 1}$

$a=b$

Thus, the inverse of the function $f\left(x\right)=x\text{ - 1}$ exists.

For the interval $\left[0,\infty \right)$, $f\left(x\right)=lnx$

Let us assume that if $f\left(a\right)=f\left(b\right)$

Then, $lna=lnb$

$a=b$

Thus, the inverse of the function $f\left(x\right)=lnx$ exists.

Hence, we get that the inverse of the function $f\left(x\right)=\left\{(x\text{ - 1) }whenx<0andlnxwhenx\ge 0\right\}$ exists.

Inverse of the function:

For the interval $(\infty ,0)$,$f\left(x\right)=x\text{ - 1}$

Replacing $f\left(x\right)$ with $y$, we get

$y=x\text{ - 1}$

On switching $y$ and $x$, we get

$x=y\text{ - 1}$

Solving for $y$, we get

$x\text{ + 1}=y$

$y=x\text{ + 1}$

Replacing $y$ with $f^{-1}\left(x\right)$, we get

$f^{-1}\left(x\right)=x\text{ + 1}$

For the interval $\left[0,\infty \right)$, $f\left(x\right)=lnx$

Replacing $f\left(x\right)$ with $y$, we get

$y=lnx$

On switching $y$ and $x$, we get

$x=lny$

Solving for $y$, we get

$e^x=e^{lny}$

$e^x=y$

$y=e^x$

Replacing $y$ with $f^{-1}\left(x\right)$, we get

$f^{-1}\left(x\right)=e^x$

Thus, Inverse of function is $f^{-1}\left(x\right)=\left\{\text{(}x\text{ + 1) }whenx<0and{e}^{x}whenx\ge 0\right\}$