Chapter 0.5, Problem 51E

(a)

To determine

To find: The convincing argument that $f$ is a one to one function.

Answer to Problem 51E

  $f\left(x_1\right)=f\left(x_2\right)$ and $x_1=x_2$ therefore by definition this function is one-to-one function.

Explanation of Solution

Given information: The function is: $y=f(x)=mx+b$

Calculation:

  $f\left(x_1\right)=m{x}_1+bf\left(x_2\right)=m{x}_2+b$

Therefore those two functions will be equal when:

  $\begin{array}{c}m{x}_1+b=m{x}_2+b\\ m{x}_1=m{x}_2\\ x_1=x_2\end{array}$

Keep in mind that we may divide by $m$ in the final step because $m\ne 0$ .

Therefore the required $f\left(x_1\right)=f\left(x_2\right)$ and $x_1=x_2$ therefore by definition this function is one-to-one function.

(b)

To determine

To find: The for the inverse of $f$ .and explain the slope $f^{-1}$ related.

Answer to Problem 51E

  $f^{-1}(x)=\frac{x}{m}-\frac{b}{m}$

The slopes are inverse.

Explanation of Solution

Given information: The function is: $y=f(x)=mx+b$

Calculation:

Switch $x$ and $y$ to solve for $y$ :

  $\begin{array}{c}x=my+b\\ my=x-b\\ y=\frac{x-b}{m}\\ y=\frac{x}{m}-\frac{b}{m}\end{array}$

The inverse function of $f$ is:

  $f^{-1}(x)=\frac{x}{m}-\frac{b}{m}$

The slope of $f^{-1}$ is $\frac{1}{m}$ .

(c)

To determine

To explain: The graph of the function is inverse.

Answer to Problem 51E

They are a pair of parallel lines as well.

Explanation of Solution

Given information: The function is: $y=f(x)=mx+b$

Explain:

Let f and g represent linear functions whose graphs are parallel lines, and thus they have the same slope $m\ne 0$ .

Then their respective inverses $f^{-1}$ and $g^{-1}$ will have the same slope $\frac{1}{m}$ . In other words, the graphs of their inverses will be a pair of parallel lines as well.

(d)

To determine

To explain: The graph of the function is inverse.

Answer to Problem 51E

They are a pair of perpendicular lines as well.

Explanation of Solution

Given information: The function is: $y=f(x)=mx+b$

Explain:

From 51(b):

  $f-1\text{ as }\frac{1}{m}$ ,the slope $f^{-1}$ as $\frac{1}{m}$ and $g^{-1}$ as ${\left(-\frac{1}{m}\right)}^{-1}=-m$

Since,

  $\frac{1}{m}(-m)=-1$

The graph of $f^{-1}$ and $g^{-1}$ are perpendicular as well.