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

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

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

Answer

Explanation:

Step1: Analyze left - hand limit

For (x < - 1), (f(x)=-x + 3). Calculate (\lim_{x\rightarrow - 1^{-}}f(x)). Substitute (x=-1) into (-x + 3), we get (\lim_{x\rightarrow - 1^{-}}(-x + 3)=-(-1)+3=4).

Step2: Analyze right - hand limit

For (x\geq - 1), (f(x)=-3x + 1). Calculate (\lim_{x\rightarrow - 1^{+}}f(x)). Substitute (x = - 1) into (-3x+1), we get (\lim_{x\rightarrow - 1^{+}}(-3x + 1)=-3\times(-1)+1=4).

Step3: Analyze function value at (x=-1)

When (x=-1), using (f(x)=-3x + 1) (since (x=-1) satisfies (x\geq - 1)), (f(-1)=-3\times(-1)+1=4).

Step4: Determine continuity

Since (\lim_{x\rightarrow - 1^{-}}f(x)=\lim_{x\rightarrow - 1^{+}}f(x)=f(-1) = 4), and the two linear - piece functions (y=-x + 3) for (x < - 1) and (y=-3x + 1) for (x\geq - 1) are continuous on their respective domains, the function (f(x)) is continuous.

Answer:

Yes