Chapter 5, Problem 13CT

(a)

To determine

The domain of $f$ , where $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Answer to Problem 13CT

Solution:

The domain of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(2,\infty \right)$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Consider the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

The given function is a logarithmic function.

The logarithmic function is defined only for positive real numbers.

The domain of $f$ consists of all $x$ for which $x-2>0$ , or equivalently, $x>2$ .

The domain of $f$ is $\left\{x|x>2\right\}$ , or equivalently, $\left(2,\infty \right)$ in interval notation.

Thus, the domain of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(2,\infty \right)$ .

(b)

To determine

To graph: The function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Graph:

Use the steps below to graph the function using graphing calculator:

Step I: Press the ON key.

Step II: Now, press [Y=]. Input the right hand side of the function $y=1-{\log }_5\left(x-2\right)$ in $Y_1$ .

Step III: Then hit [Graph] key to view the graph.

The graph of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is as follows:

  

Interpretation:

By looking at the graph, it is observed that as $x\to 2$ , $y\to \infty$ . As $x$ value increases from $x=2$ , $y$ also increases. The above graph shows that graph of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

(c)

To determine

The range of any asymptotes from the graph of $f$ , where $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Answer to Problem 13CT

Solution:

1) The range of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(-\infty ,\infty \right)$ .

2) The vertical asymptote of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $x=2$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

The domain of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(2,\infty \right)$ .

Range of the function is the set of values of the dependent variable for which a function is defined.

From the graph, range of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(-\infty ,\infty \right)$ .

The graph of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ has no horizontal asymptote.

The vertical asymptote is $x=2$ .

Therefore, the vertical asymptote of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $x=2$ .

(d)

To determine

The inverse of $f$ , where $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Answer to Problem 13CT

Solution:

The inverse of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $f^{-1}\left(x\right)=5^{-x+1}+2$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Consider the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Step 1: Replace $f\left(x\right)$ with $y$ ,

  $\Rightarrow y=1-{\log }_5\left(x-2\right)$

Step 2: Interchange the variables $x$ and $y$ .

Interchange the variables $x$ and $y$ in $y=1-{\log }_5\left(x-2\right)$ to obtain $x=1-{\log }_5\left(y-2\right)$ .

Step 3: Solve for $y$ to obtain explicit form of $f^{-1}$ .

Subtract $1$ from both sides,

  $\Rightarrow x-1=-{\log }_5\left(y-2\right)$

Multiplying both sides by $-1$ ,

  $\Rightarrow -x+1={\log }_5\left(y-2\right)$

By using, $y={\log }_{a}x$ if and only if $a^y=xa>0a\ne 1$ ,

  $\Rightarrow 5^{-x+1}=y-2$

Adding $2$ to both sides,

  $\Rightarrow y=5^{-x+1}+2$

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

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

Therefore, the inverse of $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $f^{-1}\left(x\right)=5^{-x+1}+2$ .

(e)

To determine

The domain and range of $f^{-1}$ .

Answer to Problem 13CT

Solution:

The domain of $f^{-1}\left(x\right)$ is $\left(-\infty \infty \right)$, and range of $f^{-1}\left(x\right)$ is $\left(2,\infty \right)$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

From part (d), $f^{-1}\left(x\right)=5^{-x+1}+2$ .

  $f^{-1}$ is an exponential function.

The exponential function is defined for all real numbers.

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

The domain of the function $f\left(x\right)=1-{\log }_5\left(x-2\right)$ is $\left(2,\infty \right)$ .

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

The range of $f^{-1}\left(x\right)$ is $\left(2,\infty \right)$ .

Therefore, the domain of $f^{-1}\left(x\right)$ is $\left(-\infty \infty \right)$ , and range of $f^{-1}\left(x\right)$ is $\left(2,\infty \right)$ .

(e)

To determine

To graph: The function $f^{-1}$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=1-{\log }_5\left(x-2\right)$ .

Graph:

From part (d), $f^{-1}\left(x\right)=5^{-x+1}+2$ .

Use the steps below to graph the function using graphing calculator:

Step I: Press the ON key.

Step II: Now, press [Y=]. Input the right hand side of the function $y=5^{-x+1}+2$ in $Y_1$ .

Step III: Then hit [Graph] key to view the graph.

The graph of the function $f\left(x\right)=5^{-x+1}+2$ is as follows:

  

Interpretation:

By looking at the graph, it is observed that as $x\to \infty$ , $y\to 2$ . The above graph shows that graph of the function $f^{-1}\left(x\right)=5^{-x+1}+2$ .