Question

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let (f(x)=\begin{cases}4x + 13&\text{if }x < - 2\sqrt{x + 27}&\text{if }x > - 2\\2&\text{if }x=-2end{cases}
select all statements below that you agree with.
note: you may be checking more than one box. no partial credit.
□ (f(-2)) is defined.
□ (lim_{x
ightarrow - 2}f(x)) exists.
□ (lim_{x
ightarrow - 2}f(x)=f(-2)).
□ the function is continuous at (x = - 2).
□ the function is not continuous at (x=-2).

Explanation:

Step1: Check if $f(-2)$ is defined

Given $f(-2) = 2$, so $f(-2)$ is defined.

Step2: Calculate the left - hand limit

For $x\to - 2^{-}$, $f(x)=4x + 13$. Then $\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{-}}(4x + 13)=4\times(-2)+13=5$.

Step3: Calculate the right - hand limit

For $x\to - 2^{+}$, $f(x)=\sqrt{x + 27}$. Then $\lim_{x\to - 2^{+}}f(x)=\lim_{x\to - 2^{+}}\sqrt{x + 27}=\sqrt{-2 + 27}=5$.
Since $\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{+}}f(x)=5$, $\lim_{x\to - 2}f(x)=5$.

Step4: Compare the limit and the function value at $x = - 2$

We have $\lim_{x\to - 2}f(x)=5$ and $f(-2)=2$. Since $\lim_{x\to - 2}f(x)
eq f(-2)$, the function is not continuous at $x=-2$.

Answer:

$f(-2)$ is defined.
$\lim_{x\to - 2}f(x)$ exists.
The function is not continuous at $x = - 2$.