Question

a) $lim_{x
ightarrow1^{-}}f(x)=$

b) $lim_{x
ightarrow1^{+}}f(x)=$

d) is $f$ continuous or discontinuous at $x = 1$? why?

given $f(x)=\begin{cases}2x - 4,xleq - 1\ax^{2}-3,x > - 1end{cases}$, for what value of $a$ is the function continuous?

1. find the interval(s) over which the function is continuous. be sure to show work to support your answers.

$f(x)=\frac{2}{x^{2}-9}$

Explanation:

Step1: Find the domain of the function

The function \(f(x)=\frac{2}{x^{2}-9}\) is undefined when the denominator \(x^{2}-9 = 0\). Solve \(x^{2}-9=0\), which is \((x + 3)(x - 3)=0\). So \(x=-3\) or \(x = 3\).

Step2: Determine the intervals of continuity

The function is a rational - function and is continuous on its domain. The domain is all real numbers except \(x=-3\) and \(x = 3\). So the intervals of continuity are \((-\infty,-3)\), \((-3,3)\) and \((3,\infty)\).

Answer:

The function \(f(x)=\frac{2}{x^{2}-9}\) is continuous on the intervals \((-\infty,-3)\), \((-3,3)\) and \((3,\infty)\).