Question

at what points is the following function continuous?

f(x) =
\

$$\begin{cases} \\dfrac{x^3 - 8}{x^2 - 4}, & x \ eq 2, x \ eq -2 \\\\ 3, & x = 2 \\\\ 7, & x = -2 \\end{cases}$$

choose the correct answer below and, if necessary, fill in the answer box to complete your answer.

\bigcirc a. all real numbers except at x = \boxed{\quad}
(use a comma to separate answers as needed.)

\bigcirc b. only at x = \boxed{\quad}
(use a comma to separate answers as needed.)

\bigcirc c. all real numbers x

Explanation:

Step1: Simplify the rational function

First, factor the numerator and denominator. The numerator \(x^3 - 8\) is a difference of cubes, so \(x^3 - 8=(x - 2)(x^2+2x + 4)\). The denominator \(x^2 - 4\) is a difference of squares, so \(x^2 - 4=(x - 2)(x + 2)\). Then, for \(x
eq2,x
eq - 2\), we can simplify \(\frac{x^3 - 8}{x^2 - 4}=\frac{(x - 2)(x^2+2x + 4)}{(x - 2)(x + 2)}=\frac{x^2+2x + 4}{x + 2}\) (we cancel out the non - zero factor \(x - 2\) since \(x
eq2\)).

Step2: Check continuity at \(x = 2\)

To check continuity at \(x = 2\), we need to find \(\lim_{x
ightarrow2}f(x)\). Since for \(x
eq2,x
eq - 2\), \(f(x)=\frac{x^2+2x + 4}{x + 2}\), we substitute \(x = 2\) into this simplified function: \(\lim_{x
ightarrow2}\frac{x^2+2x + 4}{x + 2}=\frac{2^2+2\times2 + 4}{2 + 2}=\frac{4 + 4+4}{4}=\frac{12}{4}=3\). And \(f(2) = 3\). So \(\lim_{x
ightarrow2}f(x)=f(2)\), which means the function is continuous at \(x = 2\).

Step3: Check continuity at \(x=-2\)

First, find \(\lim_{x
ightarrow - 2}f(x)\). For \(x
eq2,x
eq - 2\), \(f(x)=\frac{x^2+2x + 4}{x + 2}\). As \(x
ightarrow - 2\), the denominator \(x + 2
ightarrow0\), and the numerator \(x^2+2x + 4=(-2)^2+2\times(-2)+4=4-4 + 4 = 4
eq0\). So \(\lim_{x
ightarrow - 2}\frac{x^2+2x + 4}{x + 2}\) does not exist (the limit goes to \(\pm\infty\)). And \(f(-2)=7\). Since the limit as \(x
ightarrow - 2\) does not equal \(f(-2)\) (and in fact the limit does not exist), the function is discontinuous at \(x=-2\).

Step4: Check continuity for \(x

eq2,x
eq - 2\)
For \(x
eq2,x
eq - 2\), \(f(x)=\frac{x^3 - 8}{x^2 - 4}\) is a rational function, and rational functions are continuous everywhere except where the denominator is zero. Since for \(x
eq2,x
eq - 2\), the denominator \(x^2 - 4
eq0\), the function \(f(x)\) is continuous at all \(x
eq2,x
eq - 2\) (because it is a composition of continuous functions: polynomial functions are continuous everywhere, and the quotient of two continuous functions is continuous where the denominator is non - zero). And we already saw it is continuous at \(x = 2\), but discontinuous at \(x=-2\).

Answer:

A. All real numbers except at \(x=-2\)