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

Explanation:

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

The given function is a logarithmicfunction.

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

Explanation:

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, thevertical 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 $f\left(x\right)=1-{\log }_5\left(x-2\right)$.

Explanation:

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

Explanation:

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