use the function and the accompanying figure to answer the following questions.
a. is f defined at x = 2?
yes
no
b. is f continuous at x = 2?
yes
no
f(x)=\begin{cases}x^{2}-3, & -1\leq x

Question

use the function and the accompanying figure to answer the following questions.
a. is f defined at x = 2?
yes
no
b. is f continuous at x = 2?
yes
no
f(x)=\begin{cases}x^{2}-3, & -1\leq x<0 \\ 4x, & 0<x<1 \\ 6, & x = 1 \\ -3x + 7, & 1<x<2 \\ 1, & 2\leq x<3\end{cases}

Accepted Answer

# Explanation:
## Step1: Check definition at x = 2
For $2\leq x<3$, $f(x)=1$. So $f(x)$ is defined at $x = 2$.
## Step2: Check continuity at x = 2
Find left - hand limit $\lim_{x
ightarrow2^{-}}f(x)$. For $1 < x<2$, $f(x)=-3x + 7$. Then $\lim_{x
ightarrow2^{-}}(-3x + 7)=-3\times2+7=1$.
Find right - hand limit $\lim_{x
ightarrow2^{+}}f(x)$. For $2\leq x<3$, $f(x)=1$, so $\lim_{x
ightarrow2^{+}}f(x)=1$. And $f(2) = 1$. Since $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)$, the function is continuous at $x = 2$.
# Answer:
a. Yes
b. Yes

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QUESTION IMAGE

Question

Explanation:

Step1: Check definition at x = 2

Step2: Check continuity at x = 2

Answer: