Question

question 6
let $f(x)=\begin{cases}5x + 38&\text{if }x < - 6\sqrt{x + 70}&\text{if }x > - 6\\2&\text{if }x=-6end{cases}$
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 $x = - 6$, $f(-6)=2$. So $f(-6)$ is defined.

Step2: Calculate the left - hand limit

For $x\to - 6^{-}$, $f(x)=5x + 38$. Then $\lim_{x\to - 6^{-}}f(x)=\lim_{x\to - 6^{-}}(5x + 38)=5\times(-6)+38=-30 + 38 = 8$.

Step3: Calculate the right - hand limit

For $x\to - 6^{+}$, $f(x)=\sqrt{x + 70}$. Then $\lim_{x\to - 6^{+}}f(x)=\sqrt{-6 + 70}=\sqrt{64}=8$.
Since $\lim_{x\to - 6^{-}}f(x)=\lim_{x\to - 6^{+}}f(x)=8$, $\lim_{x\to - 6}f(x)=8$.

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

We have $\lim_{x\to - 6}f(x)=8$ and $f(-6)=2$. Since $\lim_{x\to - 6}f(x)
eq f(-6)$.

Step5: Determine continuity

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x\to a}f(x)=f(a)$. Since $\lim_{x\to - 6}f(x)
eq f(-6)$, 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 $x = - 6$.