suppose that the function (f) is defined, for all real numbers, as follows.\nf(x)=\begin{cases}3x +…

suppose that the function (f) is defined, for all real numbers, as follows.\nf(x)=\begin{cases}3x + 1&\text{if }x < - 2\\x - 3&\text{if }xgeq - 2end{cases}\ngraph the function (f). then determine whether or not the function is continuous.\nis the function continuous?\nyes\nno

suppose that the function (f) is defined, for all real numbers, as follows.\nf(x)=\begin{cases}3x + 1&\text{if }x < - 2\\x - 3&\text{if }xgeq - 2end{cases}\ngraph the function (f). then determine whether or not the function is continuous.\nis the function continuous?\nyes\nno

Answer

Explanation:

Step1: Find the left - hand limit as $x\to - 2$

For $x < - 2$, $f(x)=3x + 1$. Then $\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{-}}(3x + 1)=3\times(-2)+1=-6 + 1=-5$.

Step2: Find the right - hand limit as $x\to - 2$

For $x\geq - 2$, $f(x)=x - 3$. Then $\lim_{x\to - 2^{+}}f(x)=\lim_{x\to - 2^{+}}(x - 3)=-2-3=-5$.

Step3: Find the value of the function at $x = - 2$

For $x\geq - 2$, $f(-2)=-2 - 3=-5$.

Step4: Check the continuity condition

Since $\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{+}}f(x)=f(-2)=-5$, the function is continuous.

Answer:

Yes