f(x)=\\begin{cases}3x + 1 &\\text{if }x < - 2\\x - 3&\\text{if }x\\geq - 2\\end{cases}\ngraph the function…

Explanation:

Step1: Find left - hand limit

For $x\lt - 2$, $f(x)=3x + 1$. Calculate $\lim_{x\rightarrow - 2^{-}}f(x)=\lim_{x\rightarrow - 2^{-}}(3x + 1)=3\times(-2)+1=-6 + 1=-5$.

Step2: Find right - hand limit

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

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

Substitute $x = - 2$ into $f(x)=x - 3$ (since $x=-2$ satisfies $x\geq - 2$), $f(-2)=-2-3=-5$.

Step4: Check continuity condition

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

Answer:

Yes