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let $f(x)=\

$$\begin{cases}2x + 16 & \\text{if } x < -6\\\\\\sqrt{x + 22} & \\text{if } x > -6\\\\2 & \\text{if } x = -6\\end{cases}$$

$
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$\square$ $f(-6)$ is defined.
$\square$ $\lim_{x \to -6} f(x)$ exists.
$\square$ $\lim_{x \to -6} f(x)=f(-6)$.
$\square$ the function is continuous at x = -6.
$\square$ the function is not continuous at x = -6.
question help: video 1 video 2

Explanation:

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

From the piecewise function, when $x=-6$, $f(-6)=2$, so it is defined.

Step2: Calculate left-hand limit at $x=-6$

For $x<-6$, $f(x)=2x+16$. Compute:
$\lim_{x \to -6^-} f(x) = 2(-6)+16 = -12+16=4$

Step3: Calculate right-hand limit at $x=-6$

For $x>-6$, $f(x)=\sqrt{x+22}$. Compute:
$\lim_{x \to -6^+} f(x) = \sqrt{-6+22}=\sqrt{16}=4$

Step4: Verify if overall limit exists

Since $\lim_{x \to -6^-} f(x)=\lim_{x \to -6^+} f(x)=4$, $\lim_{x \to -6} f(x)=4$ exists.

Step5: Compare limit and $f(-6)$

We have $\lim_{x \to -6} f(x)=4$ and $f(-6)=2$, so $4
eq 2$.

Step6: Check continuity at $x=-6$

A function is continuous at $x=a$ if $\lim_{x \to a} f(x)=f(a)$. Since $4
eq 2$, the function is not continuous at $x=-6$.

Answer:

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