Chapter 0, Problem 69EP

(a)

To determine

To calculate : The domain of the given function.

Answer to Problem 69EP

The domain of the given function is $\left(2,\infty \right)$ .

Explanation of Solution

Given information :

The function is $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

As the domain of a function consists of all the real values of $x$ that give a real value of $y$ .

Therefore, the domain of the given function is $\left(2,\infty \right)$ .

(b)

To determine

To calculate : The range of the given function.

Answer to Problem 69EP

The range of the given function is $\left(-\infty ,\infty \right)$ or all real numbers.

Explanation of Solution

Given information :

The function is $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

As we know the range of the function is given by all real values attained by $y$ when we replace $x$ by all different values in the domain.

Therefore, the range of the given function is $\left(-\infty ,\infty \right)$ or all real numbers.

(c)

To determine

To calculate : The $x$ intercept of the graph of the given function.

Answer to Problem 69EP

The $x$ intercept of the graph of the given function is $4.718$ .

Explanation of Solution

Given information :

The function is $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

The $x$ -intercepts of the graph of the given function is:

  $\begin{array}{c}f\left(x\right)=1-\text{In}\left(x-2\right)\\ 0=1-\text{In}\left(x-2\right)\\ 1=\text{In}\left(x-2\right)\\ e^{-1}=x-2\\ x=e+2\\ x\approx 4.718\end{array}$

(d)

To determine

To calculate : The value of $f^{-1}$ of the given function.

Answer to Problem 69EP

The value of $f^{-1}$ of the given function is $2+e^{1-x}$ .

Explanation of Solution

Given information :

The function is $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

Now we get $f^{-1}$ as follows:

  $\begin{array}{l}y=1-\text{In}\left(x-2\right)\\ \Rightarrow y-1=-\text{In}\left(x-2\right)\\ \Rightarrow 1-y=\text{In}\left(x-2\right)\\ \Rightarrow e^{1-y}=x-2\end{array}$

Rearrange the above equation, to get

  $x=2+e^{1-y}$

Interchange $x\text{ and }y$ ,

  $y=2+e^{1-x}$

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

(e)

To determine

To prove : The answer in part (d) algebraically.

Explanation of Solution

Given information :

The function is $f\left(x\right)=1-\ln \left(x-2\right)$ .

Proof :

In order to prove the validity of the above equation in part (d) of the question, we find the following:

  $\begin{array}{l}\left(fo{f}^{-1}\right)\left(x\right)=f\left(f^{-1}\left(x\right)\right)\\ =f\left(2+e^{1-x}\right)\\ =1-\text{In}\left(2+e^{1-x}-2\right)\\ =1-\text{In}\left(e^{1-x}\right)\\ =1-\left(1-x\right)\\ =x\end{array}$

  $\begin{array}{l}\left(f^{-1}of\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)\\ =f^{-1}\left(1-\text{In}\left(x-2\right)\right)\\ =2+e^{1-\left(1-\text{In}\left(x-2\right)\right)}\\ =2+\left(x-2\right)\\ =x\end{array}$

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