Chapter 3.5, Problem 1QR

Solution

To prove: each function in the given pair is the inverse of the other.

Given information:

The functions:

  $f\left(x\right)=e^{2x}$ and $g\left(x\right)=\ln \left(x^{\frac{1}{2}}\right)$

Property Used:

Inverse Composition Rule

A function f is one-to-one with inverse function g if and only if $f\left(g\left(x\right)\right)=x$ for every x in the domain of g, and $g\left(f\left(x\right)\right)=x$ for every x in the domain of f .

Basic Properties of Natural Logarithms

Let x and y be real numbers with $x>0$ .

• $\ln 1=0$ because $e^0=1$ .
• $\ln e=1$ because $e^1=e$ .
• $\ln e^y=y$ because $e^y=e^y$ .
• $e^{\ln x}=x$ because $\ln x=\ln x$

Explanation:

The domain of the function $f\left(x\right)=e^{2x}$ is $\left(-\infty ,\infty \right)$ which is the range of function $g\left(x\right)=\ln \left(x^{\frac{1}{2}}\right)$ .

And the domain of the function $g\left(x\right)=\ln \left(x^{\frac{1}{2}}\right)$ is $\left(0,\infty \right)$ which is the range of function $f\left(x\right)=e^{2x}$ .

Using inverse composition rule,

  $\begin{array}{c}f\left(g\left(x\right)\right)=e^{2g\left(x\right)}\\ =e^{2\times \frac{1}{2}\ln x}\\ =e^{\ln x}\\ =x\end{array}$

  $\begin{array}{c}g\left(f\left(x\right)\right)=\frac{1}{2}\ln \left(f\left(x\right)\right)\\ =\frac{1}{2}\ln \left(e^{2x}\right)\\ =\frac{1}{2}\times 2\ln e^x\\ =x\end{array}$

Since these equations are true for all x , the Inverse Composition Rule guarantees that f and g are inverses.