Q 82. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 71– 86, use the given function f  to:

(a) Find the domain of f.

(b) Graph f.

(c) From the graph, determine the range and any asymptotes of f.

(d) Find $f^{-1}$, the inverse of f.

(e) Find the domain and the range of $f^{-1}$

(f) Graph $f^{-1}$

$f\left(x\right)=2-{\log }_3(x+1)$

Step-by-Step Solution

Answer

Part $a=\left(-1,\infty \right)$

Part b =

Part (c) Range $\left(-\infty ,\infty \right)$ and vertical asymptote is $x=-1$

Part (d) $f^{-1}\left(x\right)=3^{2-x}-1$.

Part (e) Domain $\left(-\infty ,\infty \right)$ and range$\left(-1,\infty \right)$.

Part (f) 

Part (a) Step 1. Given information.

The given function is: $f\left(x\right)=2-{\log }_3(x+1)$

Part (a) Step 2. Find the domain of f.

The domain of f consists of all x for which $\left(x+1\right)>0$.

Therefore, the domain of the given function is $\left(-1,\infty \right)$

Part (b) Step 1. Graph f.

Sketch the graph of f : 

Part (c) Step 1. Determine the range and any asymptotes of f from the graph.

We can see from the graph of f that the range of the function $f\left(x\right)=2-{\log }_3(x+1)$ is the set of all real numbers. 

Therefore, the range of the function is $\left(-\infty ,\infty \right)$

The vertical asymptote of the given function is  $x=-1$

Part (d) Step 1. Find f - 1 .

$f\left(x\right)=2-{\log }_3(x+1)$

For $f^{-1}$ replace $f\left(x\right) with y$.

$$\begin{gathered} x=2-{\log }_3(x+1) \\ {\log }_3(x+1)=2-x \\ y+1=3^{2-x} \\ y=3^{2-x}-1 \end{gathered}$$

Therefore, the inverse of f  is $f^{-1}\left(x\right)=3^{2-x}-1$

Part (e) Find the domain and the range of f - 1 .

We know that the domain of a function f(x) is the range of its inverse.

In the same way, the range of a function f(x) is the domain of its inverse.

Therefore, the domain of $f^{-1} is \left(-\infty ,\infty \right)$and its range is $\left(-1,\infty \right)$

Part (f) Graph f - 1 .

Sketch the graph of $f^{-1}$.