Chapter 1, Problem 69RE

a.

To determine

To find: the domain of f.

Answer to Problem 69RE

The domain is $(2,\infty )$ .

Explanation of Solution

Given information: Let $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

The log of a negative number is undefined so $x-2>0\Rightarrow x>2$ ,so the domain is $(2,\infty )$ .

b.

To determine

To find: the range of f.

Answer to Problem 69RE

The range is $(-\infty ,\infty )$ .

Explanation of Solution

Given information: $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

All logarithmic functions have a range of all real numbers so the range is $(-\infty ,\infty )$ .

c.

To determine

To find: arc the x-intercepts of the graph of f.

Answer to Problem 69RE

The x − intercepts is $(2+e,0)\approx (4.718,0)$ .

Explanation of Solution

Given information: $f\left(x\right)=1-\ln \left(x-2\right)$

Calculation:

The x − intercepts are when y = 0, so:

  $\begin{array}{c}0=1-\ln \left(x-2\right)\\ \ln (x-2)=1\\ e^{\ln (x-2)}=e^1\\ x-2=e^1\\ x-2=e\\ x=e+2.\end{array}$

The x − intercepts is $(2+e,0)\approx (4.718,0)$ .

d.

To determine

To find: $f^{-1}$ .

Answer to Problem 69RE

  $f^{-1}(x)=e^{1-x}+2.$

Explanation of Solution

Given information: $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

To find the inverse, switch x and y and then solve for y:

  $\begin{array}{l}x=1-\ln (y-2)\\ x-1=-\ln (y-2)\\ 1-x=\ln (y-2)\\ \ln (y-2)=1-x\\ e^{\ln (y-2)}=e^{1-x}\\ y-2=e^{1-x}\\ y=e^{1-x}+2\\ f^{-1}(x)=e^{1-x}+2.\end{array}$

e.

To determine

To confirm: answer algebraically in part (d).

Answer to Problem 69RE

Explanation of Solution

Given information: $f\left(x\right)=1-\ln \left(x-2\right)$ .

Calculation:

To functions are inverses if $f\circ f^{-1}=f^{-1}\circ f=x$ .

  $\begin{array}{c}f\circ f^{-1}=f(e^{1-x}+2)\\ =1-\ln (e^{1-x}+2-2)\\ =1-(1-x)\\ =x\end{array}$

  $\begin{array}{c}{f}^{-1}\circ f=f^{-1}(1-\ln (x-2))\\ =e^{1-(1-\ln (x-2))}+2\\ =e^{\ln (x-2)}+2\\ =x-2+2\\ =x\end{array}$

Therefore, $f(x)=1-\ln (x-2)and{f}^{-1}(x)=e^{1-x}+2$ are inverses.