suppose that the function (f) is defined, for all real numbers, as follows.\nf(x)=\begin{cases}2x…

suppose that the function (f) is defined, for all real numbers, as follows.\nf(x)=\begin{cases}2x - 1&\text{if }x<1\\-x + 2&\text{if }xgeq1end{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}2x - 1&\text{if }x<1\\-x + 2&\text{if }xgeq1end{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 approaches 1

For (x < 1), (f(x)=2x - 1). (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}(2x - 1)=2\times1-1 = 1)

Step2: Find the right - hand limit as x approaches 1

For (x\geq1), (f(x)=-x + 2). (\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(-x + 2)=-1 + 2=1)

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

For (x = 1), (f(x)=-x + 2), so (f(1)=-1 + 2=1)

Step4: Check the continuity condition

Since (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{+}}f(x)=f(1) = 1), the function is continuous.

Answer:

Yes