Chapter 1.7, Problem 61PFA

(a)

To determine

To find:The value of the given functions.

Explanation of Solution

Consider the given function.

  $f\left(x\right)=x^2+2x-5$

The value of $f\left(-3\right)$ is calculated as,

  $\begin{array}{c}f\left(-3\right)={\left(-3\right)}^2+2\left(-3\right)-5\\ =9-6-5\\ =-2\end{array}$

The value of $f\left(0\right)$ is calculated as,

  $\begin{array}{c}f\left(0\right)={\left(0\right)}^2+2\left(0\right)-5\\ =0+0-5\\ =-5\end{array}$

The value of $f\left(1\right)$ is calculated as,

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^2+2\left(1\right)-5\\ =1+2-5\\ =-2\end{array}$

The value of $f\left(-a\right)$ is calculated as,

  $\begin{array}{c}f\left(-a\right)={\left(-a\right)}^2+2\left(-a\right)-5\\ =a^2-2a-5\end{array}$

(b)

To determine

To find:Whether the relation is function.

Explanation of Solution

Each value of $x$ is paired with exactly one value of $f\left(x\right)$ . So the function is a relation.

Therefore, the function is a relation.

(c)

To determine

To find:The relation is discrete or continuous.

Explanation of Solution

Consider the given function.

  $f\left(x\right)=x^2+2x-5$

The value of $f\left(-3\right)$ is calculated as,

  $\begin{array}{c}f\left(-3\right)={\left(-3\right)}^2+2\left(-3\right)-5\\ =9-6-5\\ =-2\end{array}$

The value of $f\left(0\right)$ is calculated as,

  $\begin{array}{c}f\left(0\right)={\left(0\right)}^2+2\left(0\right)-5\\ =0+0-5\\ =-5\end{array}$

The value of $f\left(1\right)$ is calculated as,

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^2+2\left(1\right)-5\\ =1+2-5\\ =-2\end{array}$

The value of $f\left(-a\right)$ is calculated as,

  $\begin{array}{c}f\left(-a\right)={\left(-a\right)}^2+2\left(-a\right)-5\\ =a^2-2a-5\end{array}$

Since,

  $f\left(-3\right)=f\left(1\right)=-2$

So, the relation is discrete.