Question

explain why the function is discontinuous at the given number a. (select all that apply.)

$f(x) = \frac{1}{x + 4} \quad a = -4$

$\square \\, f(-4)$ is undefined.
$\square \\, \lim_{x \to -4^+} f(x)$ and $\lim_{x \to -4^-} f(x)$ are not finite, and are not equal.
$\square \\, f(-4)$ and $\lim_{x \to -4} f(x)$ are finite, but are not equal.
$\square \\, \lim_{x \to -4} f(x)$ does not exist.
$\square \\, $none of the above

sketch the graph of the function.

Explanation:

Step1: Analyze \( f(-4) \)

For the function \( f(x)=\frac{1}{x + 4} \), when \( x=-4 \), the denominator \( x + 4=-4 + 4 = 0 \). Division by zero is undefined, so \( f(-4) \) is undefined.

Step2: Analyze the limit as \( x

ightarrow - 4 \)

• As \( x

ightarrow - 4^{+} \) (approaching -4 from the right), \( x+4
ightarrow0^{+} \), so \( f(x)=\frac{1}{x + 4}
ightarrow+\infty \).

• As \( x

ightarrow - 4^{-} \) (approaching -4 from the left), \( x + 4
ightarrow0^{-} \), so \( f(x)=\frac{1}{x + 4}
ightarrow-\infty \).

Since the left - hand limit \( \lim_{x
ightarrow - 4^{-}}f(x)=-\infty \) and the right - hand limit \( \lim_{x
ightarrow - 4^{+}}f(x)=+\infty \), the two one - sided limits are not finite and not equal. Also, since the one - sided limits are not equal (and not finite), \( \lim_{x
ightarrow - 4}f(x) \) does not exist. And we already know that \( f(-4) \) is undefined.

The option " \( f(-4) \) and \( \lim_{x
ightarrow - 4}f(x) \) are finite, but are not equal" is wrong because \( f(-4) \) is undefined and the limit is not finite.

Answer:

• \( f(-4) \) is undefined.
• \( \lim_{x

ightarrow - 4^{+}}f(x) \) and \( \lim_{x
ightarrow - 4^{-}}f(x) \) are not finite, and are not equal.

• \( \lim_{x

ightarrow - 4}f(x) \) does not exist.