a) $lim_{x\rightarrow1^{-}}f(x)=$\n\nb) $lim_{x\rightarrow1^{+}}f(x)=$\n\nd) is $f$ continuous or…

a) $lim_{x\rightarrow1^{-}}f(x)=$\n\nb) $lim_{x\rightarrow1^{+}}f(x)=$\n\nd) is $f$ continuous or discontinuous at $x = 1$? why?\n\ngiven $f(x)=\begin{cases}2x - 4,xleq - 1\\ax^{2}-3,x > - 1end{cases}$, for what value of $a$ is the function continuous?\n\n3. find the interval(s) over which the function is continuous. be sure to show work to support your answers.\n\n$f(x)=\frac{2}{x^{2}-9}$

a) $lim_{x\rightarrow1^{-}}f(x)=$\n\nb) $lim_{x\rightarrow1^{+}}f(x)=$\n\nd) is $f$ continuous or discontinuous at $x = 1$? why?\n\ngiven $f(x)=\begin{cases}2x - 4,xleq - 1\\ax^{2}-3,x > - 1end{cases}$, for what value of $a$ is the function continuous?\n\n3. find the interval(s) over which the function is continuous. be sure to show work to support your answers.\n\n$f(x)=\frac{2}{x^{2}-9}$

Answer

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)).