Chapter 1.7, Problem 66PPE

(a)

To determine

For the function $f\left(x\right)=x^2+2x-5$ . Find the value of $f\left(-3\right),f\left(0\right),f\left(1\right),f\left(-a\right)$ .

Answer to Problem 66PPE

  $\begin{array}{l}f\left(-3\right)=-2\\ f\left(0\right)=-5\\ f\left(1\right)=-2\\ f\left(-a\right)=a^2-2a-5\end{array}$

Explanation of Solution

Given $f\left(x\right)=x^2+2x-5.$

  $\begin{array}{l}f\left(-3\right)={\left(-3\right)}^2+2\times -3-5=-2.\\ f\left(0\right)={\left(0\right)}^2+2\times 0-5=-5.\\ f\left(1\right)={\left(1\right)}^2+2\times 1-5=-2.\\ f\left(-a\right)={\left(-a\right)}^2+2\times \left(-a\right)-5=a^2-2a-5\end{array}$

(b)

To determine

Is the relation a function?

Answer to Problem 66PPE

No the given relation is not a function.

Explanation of Solution

A relation is said to be a function if for all value of $x$ there is only one value of $y$ .

From part (a) it is clear that for different value of $x$ there is different value of $y$ .

Therefore, the given relation is not a function.

(c)

To determine

Is the relation discrete or continuous?

Answer to Problem 66PPE

The relation is continuous.

Explanation of Solution

As the value of domain and range are not countable or definite that is why this relation is continuous.