Chapter 0.2, Problem 43E

To determine

To Write:

A piecewise-defined function for the given graph.

Answer to Problem 43E

  $f\left(x\right)=\left\{\begin{array}{ll}2-x,\hfill & 0<x\le 2\hfill \\ \frac{8}{3}-\frac{x}{3},\hfill & 2<x\le 5\hfill \end{array}\right.$ .

Explanation of Solution

Given:

The graph:

  

Concepts Used:

Reading the graph of a piecewise function.

The equation of the graph of a line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Isolating a variable from and equation.

Calculations:

The given graph shows a curve having two pieces.

The first piece is in the interval $0<x\le 2$ . It is a line passing through the points $\left(0,2\right)$ and $\left(2,0\right)$ . Write the equation of this line and isolate the variable $y$ from it.

  $\begin{array}{c}\frac{y-2}{x-0}=\frac{0-2}{2-0}\\ \frac{y-2}{x}=-1\\ y-2=-x\\ y=2-x\end{array}$

The second piece is in the interval $2<x\le 5$ . It is a line passing through the points $\left(2,1\right)$ and $\left(5,0\right)$ . Write the equation of this line and isolate the variable $y$ from it.

  $\begin{array}{c}\frac{y-1}{x-2}=\frac{0-1}{5-2}\\ \frac{y-1}{x-2}=-\frac{1}{3}\\ y-1=-\frac{1}{3}\left(x-2\right)\\ y=1-\frac{1}{3}\left(x-2\right)\\ y=2-\frac{x}{3}+\frac{2}{3}\\ y=\frac{8}{3}-\frac{x}{3}\end{array}$

Combine the two pieces to get the piecewise defined function.

  $f\left(x\right)=\left\{\begin{array}{ll}2-x,\hfill & 0<x\le 2\hfill \\ \frac{8}{3}-\frac{x}{3},\hfill & 2<x\le 5\hfill \end{array}\right.$

Conclusion:

The piecewise defined function for the given graph is $f\left(x\right)=\left\{\begin{array}{ll}2-x,\hfill & 0<x\le 2\hfill \\ \frac{8}{3}-\frac{x}{3},\hfill & 2<x\le 5\hfill \end{array}\right.$ .