Question

question 5
let $f(x)=\

$$\begin{cases}4x + 11 & \\text{if } x < -2 \\\\ \\sqrt{x + 11} & \\text{if } x > -2 \\\\ 2 & \\text{if } x = -2\\end{cases}$$

$
select all statements below that you agree with.
note: you may be checking more than one box. no partial credit.
$\square$ $f(-2)$ is defined.
$\square$ $\lim_{x\to -2} f(x)$ exists.
$\square$ $\lim_{x\to -2} f(x)=f(-2)$.
$\square$ the function is continuous at x = -2.
$\square$ the function is not continuous at x = -2.
question help: video 1 video 2

Explanation:

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

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

Step2: Calculate left-hand limit at $x\to-2$

For $x<-2$, $f(x)=4x+11$.
$\lim_{x\to-2^-} f(x) = 4(-2)+11 = -8+11=3$

Step3: Calculate right-hand limit at $x\to-2$

For $x>-2$, $f(x)=\sqrt{x+11}$.
$\lim_{x\to-2^+} f(x) = \sqrt{-2+11}=\sqrt{9}=3$

Step4: Check if overall limit exists

Since left-hand limit = right-hand limit = 3, $\lim_{x\to-2} f(x)=3$ exists.

Step5: Compare limit and $f(-2)$

$\lim_{x\to-2} f(x)=3$, $f(-2)=2$. So $3
eq2$, meaning $\lim_{x\to-2} f(x)
eq f(-2)$.

Step6: Check continuity at $x=-2$

A function is continuous at $x=a$ if $\lim_{x\to a} f(x)=f(a)$. Since this is not true here, 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.