Chapter 2.3, Problem 25E

To determine

To find: The formula for the extended function $f\left(x\right)=\frac{x^2-9}{x+3}$ such that $f\left(x\right)$ is continuous at $x=-3$ .

Answer to Problem 25E

The formula for the extended function $f\left(x\right)=\frac{x^2-9}{x+3}$ such that $f\left(x\right)$ is continuous at $x=-3$ is $y=x-3$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{x^2-9}{x+3}$ and the point is $x=-3$ .

Calculation:

The value of the function $f\left(x\right)=\frac{x^2-9}{x+3}$ at $x=-3$ doesn’t exist.

Simplify the given function.

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

The function $y=x-3$ is a polynomial function and it is continuous for all the values of $x$ .

Therefore, the formula for the extended function $f\left(x\right)=\frac{x^2-9}{x+3}$ such that $f\left(x\right)$ is continuous at $x=-3$ is $y=x-3$ .