Question

let $f(x)=\begin{cases}2x + 16&\text{if }x < - 6\sqrt{x + 22}&\text{if }x>-6\\2&\text{if }x = - 6end{cases}$
select all statements below that you agree with.
note: you may be checking more than one box. no partial credit.
$square f(-6)$ is defined.
$squarelim_{x
ightarrow - 6}f(x)$ exists.
$squarelim_{x
ightarrow - 6}f(x)=f(-6)$.
$square$ the function is continuous at $x=-6$.
$square$ the function is not continuous at $x=-6$.

Explanation:

Step1: Check if $f(-6)$ is defined

When $x = - 6$, $f(x)=2$. So $f(-6)$ is defined.

Step2: Calculate the left - hand limit

For $x\lt - 6$, $f(x)=2x + 16$. Then $\lim_{x
ightarrow - 6^{-}}f(x)=\lim_{x
ightarrow - 6^{-}}(2x + 16)=2\times(-6)+16 = 4$.

Step3: Calculate the right - hand limit

For $x\gt - 6$, $f(x)=\sqrt{x + 22}$. Then $\lim_{x
ightarrow - 6^{+}}f(x)=\lim_{x
ightarrow - 6^{+}}\sqrt{x + 22}=\sqrt{-6 + 22}=\sqrt{16}=4$.
Since $\lim_{x
ightarrow - 6^{-}}f(x)=\lim_{x
ightarrow - 6^{+}}f(x)=4$, $\lim_{x
ightarrow - 6}f(x)=4$.

Step4: Compare the limit and the function value at $x=-6$

We have $\lim_{x
ightarrow - 6}f(x)=4$ and $f(-6)=2$. Since $\lim_{x
ightarrow - 6}f(x)
eq f(-6)$, the function is not continuous at $x = - 6$.

Answer:

$f(-6)$ is defined.
$\lim_{x
ightarrow - 6}f(x)$ exists.
The function is not continuous at $x=-6$.