Chapter 1.5, Problem 62E

(a)

To determine

To explain: An argument to show that the function $f\left(x\right)=\frac{ax+b}{cx+d}$ is one-to-one function.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ such that $c\ne 0$ and $ad-bc\ne 0$ .

Assume that $f\left(x_1\right)=f\left(x_2\right)$ for the given function. So,

  $\begin{array}{c}\frac{a{x}_1+b}{c{x}_1+d}=\frac{a{x}_2+b}{c{x}_2+d}\\ \left(a{x}_1+b\right)\left(c{x}_2+d\right)=\left(a{x}_2+b\right)\left(c{x}_1+d\right)\\ ac{x}_1{x}_2+ad{x}_1+bc{x}_2+bd=ac{x}_1{x}_2+ad{x}_2+bc{x}_1+bd\\ ad{x}_1+bc{x}_2=ad{x}_2+bc{x}_1\end{array}$

Simplify further.

  $\begin{array}{c}ad{x}_1-bc{x}_1=ad{x}_2-bc{x}_2\\ \left(ad-bc\right)x_1=\left(ad-bc\right)x_2\end{array}$

It is given that $ad-bc\ne 0$ . So, $x_1=x_2$ .

Therefore, it can be said that the given function is one-to-one as $f\left(x_1\right)=f\left(x_2\right)$ if and only if $x_1=x_2$ .

(b)

To determine

To find: The formula for the inverse of the given function $f\left(x\right)=\frac{ax+b}{cx+d}$ .

Answer to Problem 62E

The formula for the inverse of the given function is $f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ such that $c\ne 0$ and $ad-bc\ne 0$ .

Calculation:

Simplify the function for x in terms of y .

  $\begin{array}{c}y=\frac{ax+b}{cx+d}\\ cxy+dy=ax+b\\ cxy-ax=b-dy\\ x=\frac{b-dy}{cy-a}\end{array}$

Interchange the terms x and y to find the inverse function.

  $\begin{array}{c}y=\frac{-dx+b}{cx-a}\\ f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}\end{array}$

Therefore, the formula for the inverse of the given function is $f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$ .

(c)

To determine

To find: The horizontal and vertical asymptotes of the given function $f\left(x\right)=\frac{ax+b}{cx+d}$ .

Answer to Problem 62E

The horizontal asymptote of the function is $y=\frac{a}{c}$ and vertical asymptote is $x=-\frac{d}{c}$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ such that $c\ne 0$ and $ad-bc\ne 0$ .

Calculation:

Simplify the given function.

  $\begin{array}{c}f\left(x\right)=\frac{ax+b}{cx+d}\\ =\frac{x\left(a+\frac{b}{x}\right)}{x\left(c+\frac{d}{x}\right)}\\ =\frac{\left(a+\frac{b}{x}\right)}{\left(c+\frac{d}{x}\right)}\end{array}$

As $x\to \pm \infty$ , the function approaches to $\frac{a}{c}$ . So, the horizontal asymptote is $y=\frac{a}{c}$ such that $c\ne 0$ .

It can be seen that the function $f\left(x\right)=\frac{ax+b}{cx+d}$ is undefined at the point $x=-\frac{d}{c}$ as the denominator becomes 0. So, the vertical asymptote is $x=-\frac{d}{c}$ .

Therefore, the horizontal asymptote of the function is $y=\frac{a}{c}$ and vertical asymptote is $x=-\frac{d}{c}$ .

(d)

To determine

To find: The horizontal and vertical asymptotes of the given function $f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$ .

Answer to Problem 62E

The horizontal asymptote of the function is $y=-\frac{d}{c}$ and vertical asymptote is $x=\frac{a}{c}$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ such that $c\ne 0$ and $ad-bc\ne 0$ .

Calculation:

From part (b), the inverse function is $f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$ .

Simplify the given function.

  $\begin{array}{c}{f}^{-1}\left(x\right)=\frac{-dx+b}{cx-a}\\ =\frac{x\left(-d+\frac{b}{x}\right)}{x\left(c-\frac{a}{x}\right)}\\ =\frac{\left(-d+\frac{b}{x}\right)}{\left(c-\frac{a}{x}\right)}\end{array}$

As $x\to \pm \infty$ , the function approaches to $-\frac{d}{c}$ . So, the horizontal asymptote is $y=-\frac{d}{c}$ such that $c\ne 0$ .

It can be seen that the function $f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$ is undefined at the point $x=\frac{a}{c}$ as the denominator becomes 0. So, the vertical asymptote is $x=\frac{a}{c}$ .

Therefore, the horizontal asymptote of the function is $y=-\frac{d}{c}$ and vertical asymptote is $x=\frac{a}{c}$ .