Chapter 5, Problem 3CT

To determine

The inverse function of $f\left(x\right)=\frac{2}{3x-5}$ , and verify the answer as well. Also state the domain and range of $f$ and $f^{-1}$ .

Answer to Problem 3CT

Solution:

  $f^{-1}\left(x\right)=\frac{2+5x}{3x}$.

Domain of $f(x)$ is $\left(-\infty ,\frac{5}{3}\right)\cup \left(\frac{5}{3},\infty \right)$ , and range of $f(x)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Domain of $f^{-1}\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$, and range of $f^{-1}\left(x\right)$ is $\left(-\infty ,\frac{5}{3}\right)\cup \left(\frac{5}{3},\infty \right)$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{2}{3x-5}$ .

Step 1: Replace $f\left(x\right)$ with $y$ and interchange the variables $x$ and $y$ in $y=c$ to obtain $x=\frac{2}{3y-5}$ .

Step 2: Solve $x=\frac{2}{3y-5}$ for $y$ to obtain explicit form of $f^{-1}$ .

Multiplying both sides by $3y-5$ ,

  $\Rightarrow x\left(3y-5\right)=\frac{2}{3y-5}\left(3y-5\right)$

  $\Rightarrow x\left(3y-5\right)=2$

Dividing both sides by $x$ ,

  $\Rightarrow \frac{x\left(3y-5\right)}{x}=\frac{2}{x}$

  $\Rightarrow 3y-5=\frac{2}{x}$

Adding $5$ to both sides,

  $\Rightarrow 3y-5+5=\frac{2}{x}+5$

  $\Rightarrow 3y=\frac{2}{x}+5$

Dividing both sides by $3$ ,

  $\Rightarrow \frac{3y}{3}=\frac{\frac{2}{x}}{3}+\frac{5}{3}$

  $\Rightarrow y=\frac{\frac{2}{x}+5}{3}$

  $\Rightarrow y=\frac{\frac{2+5x}{x}}{3}$

  $\Rightarrow y=\frac{2+5x}{3x}$

Replace $y$ by $f^{-1}\left(x\right)$ ,

  $\Rightarrow f^{-1}\left(x\right)=\frac{2+5x}{3x}$

Step 3: Check the result by showing that $f^{-1}\left(f\left(x\right)\right)=x=f\left(f^{-1}\left(x\right)\right)$ .

  $f^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\frac{2}{3x-5}\right)=\frac{2+5\left(\frac{2}{3x-5}\right)}{3\left(\frac{2}{3x-5}\right)}$

  $=\frac{\left(2+\frac{10}{3x-5}\right)\left(3x-5\right)}{6}$

  $=\frac{2\left(3x-5\right)+10}{6}$

  $\frac{6x-10+10}{6}$

  $\Rightarrow \frac{6x}{6}=x$

  $f\left(f^{-1}\left(x\right)\right)=f\left(\frac{2+5x}{3x}\right)=\frac{2}{3\left(\frac{2+5x}{3x}\right)-5}$

  $=\frac{2}{\frac{2+5x}{x}-5}$

  $=\frac{2}{\frac{2+5x}{x}-\frac{5x}{x}}$

  $=\frac{2}{\frac{2+5x-5x}{x}}$

  $=\frac{2x}{2}$

  $=x$

The inverse function $f^{-1}$ of $f\left(x\right)=\frac{2}{3x-5}$ is $f^{-1}\left(x\right)=\frac{2+5x}{3x}$ .

Now, find the domain and the range of $f$ and $f^{-1}$ .

The function is $f\left(x\right)=\frac{2}{3x-5}$ .

Take the denominator of $\frac{2}{3x-5}$ , and compare it to zero to find the undefined points.

  $\Rightarrow 3x-5=0$

Add $5$ to both sides,

  $\Rightarrow 3x-5+5=0+5$

  $\Rightarrow 3x=5$

Divide both sides by $3$ ,

  $\Rightarrow \frac{3x}{3}=\frac{5}{3}$

  $\Rightarrow x=\frac{5}{3}$

  $\Rightarrow x<\frac{5}{3}$ or $x>\frac{5}{3}$

Thus, the domain of $f\left(x\right)$ is $\left(-\infty ,\frac{5}{3}\right)\cup \left(\frac{5}{3},\infty \right)$.

By using domain of $f\left(x\right)$ is range of $f^{-1}\left(x\right)$ ,

Range of $f^{-1}\left(x\right)$ is $\left(-\infty ,\frac{5}{3}\right)\cup \left(\frac{5}{3},\infty \right)$ .

The range of $f\left(x\right)=\frac{2}{3x-5}$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

By using range of $f\left(x\right)$ is equal to domain of $f^{-1}\left(x\right)$ ,

The domain of $f^{-1}\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

The domain of $f\left(x\right)$ is $\left(-\infty ,\frac{5}{3}\right)\cup \left(\frac{5}{3},\infty \right)$

  $=$ range of $f^{-1}\left(x\right)$ , and range of $f\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

  $=$ domain of $f^{-1}\left(x\right)$ .