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 < - 1\\-x + 1&\text{if }xgeq - 1end{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 < - 1\\-x + 1&\text{if }xgeq - 1end{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 $x\to - 1$

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

Step2: Find the right - hand limit as $x\to - 1$

For $x\geq - 1$, $f(x)=-x + 1$. Then $\lim_{x\to - 1^{+}}f(x)=\lim_{x\to - 1^{+}}(-x + 1)=-(-1)+1=1 + 1 = 2$.

Step3: Find the value of the function at $x=-1$

For $x=-1$, using $f(x)=-x + 1$ (since $x=-1$ satisfies $x\geq - 1$), $f(-1)=-(-1)+1=2$.

Step4: Check the continuity condition

Since $\lim_{x\to - 1^{-}}f(x)=\lim_{x\to - 1^{+}}f(x)=f(-1) = 2$, the function is continuous.

Answer:

Yes