Chapter 3.4, Problem 48E

Solution

To graph: The function $f\left(x\right)={\log }_{\frac{1}{3}}\left(9x\right)$ and analyze it for domain, range, continuity, increasing or decreasing behavior, asymptotes, and end behavior.

Given information:

The function $f\left(x\right)={\log }_{\frac{1}{3}}\left(9x\right)$ .

Graph:

The given function is-

  $f\left(x\right)={\log }_{\frac{1}{3}}\left(9x\right)$

Substitute the few values of $x$ in function $f\left(x\right)={\log }_{\frac{1}{3}}\left(9x\right)$ ,

For $x=0.00001,\frac{1}{9},\frac{1}{3},1,9$:

  $\begin{array}{l}f\left(0.00001\right)={\log }_{\frac{1}{3}}\left(9\times 0.00001\right)\\ ={\log }_{\frac{1}{3}}\left(0.00009\right)\\ =8.5\end{array}$

  $\begin{array}{l}f\left(\frac{1}{9}\right)={\log }_{\frac{1}{3}}\left(9\times \frac{1}{9}\right)\\ ={\log }_{\frac{1}{3}}\left(1\right)\\ =0\end{array}$

  $\begin{array}{l}f\left(\frac{1}{3}\right)={\log }_{\frac{1}{3}}\left(9\times \frac{1}{3}\right)\\ ={\log }_{\frac{1}{3}}\left(3\right)\\ =-1\end{array}$

  $\begin{array}{l}f\left(1\right)={\log }_{\frac{1}{3}}\left(9\times 1\right)\\ ={\log }_{\frac{1}{3}}\left(9\right)\\ =-2\end{array}$

  $\begin{array}{l}f\left(9\right)={\log }_{\frac{1}{3}}\left(9\times 9\right)\\ ={\log }_{\frac{1}{3}}\left(81\right)\\ =-4\end{array}$

  

Interpretation:

Domain: $\left(0,\infty \right)$

Range: $\left(-\infty ,\infty \right)$

Continuity: Function is non-continuous.

Increasing or decreasing behavior: Function is decreasing on its domain.

Asymptotes: Function has one vertical asymptote at $x=0$ .

End behavior: $\underset{x\to \infty }{\lim }{\log }_{\frac{1}{3}}\left(9x\right)=-\infty$