Chapter 0.5, Problem 62E

a.

To determine

The given function $f$ is one to one by using the convincing argument.

Answer to Problem 62E

By the horizontal line test, the function is one to one function.

Explanation of Solution

Given Information:

The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ , Where, $c\ne 0$ , $ad-bc\ne 0$ .

Consider the given function,

  $f\left(x\right)=\frac{ax+b}{cx+d}$

It is given that $ad-bc\ne 0$ . The value of $ad$ and $bc$ cannot be equal. So there will be a unique value of $f\left(x\right)$ for each $x$ . This implies that the graph of the function will not intersect the horizontal line more than one time. The horizontal line test will hold.

Which is indicating that the given function is one to one function.

Therefore, the function is one to one function.

b.

To determine

To calculate: The inverse formula of the given function $f$ .

Answer to Problem 62E

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

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ .

Calculation:

Consider the given equation,

  $f\left(x\right)=\frac{ax+b}{cx+d}$

Replace $x$ with $y=f\left(x\right)$ to each other.

  $\begin{array}{c}x=\frac{ay+b}{cy+d}\\ cxy+xd=ay+b\\ xd-b=ay-cxy\end{array}$

Further simplify.

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

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

c.

To determine

To calculate: The vertical and horizontal asymptote of the given function $f$ .

Answer to Problem 62E

The vertical and horizontal asymptote is $\frac{-d}{c}$ and $\frac{a}{c}$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{ax+b}{cx+d}$ .

Calculation:

Consider the given equation,

Equate the denominator to find the vertical asymptote of the function.

  $\begin{array}{c}cx+d=0\\ cx=-d\\ x=\frac{-d}{c}\end{array}$

And, to find the horizontal asymptote, take the limit $x$ tends to infinite for the given function.

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

Therefore, the vertical and horizontal asymptote is $\frac{-d}{c}$ and $\frac{a}{c}$ .

d.

To determine

To calculate: The vertical and horizontal asymptote of the given function $f$ .

Answer to Problem 62E

The asymptotes are equal.

Explanation of Solution

Given information:

The inverse function is $f^{-1}\left(x\right)=\frac{xd-b}{a-cx}$ .

Calculation:

Consider the given function,

  $f^{-1}\left(x\right)=\frac{xd-b}{a-cx}$

Equate the denominator to find the vertical asymptote of the function.

  $\begin{array}{c}a-cx=0\\ a=cx\\ x=\frac{a}{c}\end{array}$

And, to find the horizontal asymptote, take the limit $x$ tends to infinite for the given function.

  $\begin{array}{c}\underset{x\to \infty }{\lim }{f}^{-1}\left(x\right)=\underset{x\to \infty }{\lim }\frac{xd-b}{a-cx}\\ =\underset{x\to \infty }{\lim }\frac{x\left(d-\frac{b}{x}\right)}{x\left(\frac{a}{x}-c\right)}\\ =\frac{\left(d-\frac{b}{\infty }\right)}{\left(\frac{a}{\infty }-c\right)}\\ =-\frac{d}{c}\end{array}$

So the asymptote of the function $f$ and inverse function of $f$ are equal.

Therefore, the vertical and horizontal asymptote is $\frac{-d}{c}$ and $\frac{a}{c}$ .