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

suppose that the function f is defined, for all real numbers, as follows.\nf(x)=\begin{cases}-x + 2&\text{if }xleq1\\5x - 2&\text{if }x>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 approaches 1

We use the part of the function for (x\leq1), so (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}(-x + 2)). Substitute (x = 1) into (-x+2), we get (-1 + 2=1).

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

We use the part of the function for (x>1), so (\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(5x - 2)). Substitute (x = 1) into (5x-2), we get (5\times1-2 = 3).

Step3: Evaluate the function at x = 1

We use the part of the function for (x\leq1), so (f(1)=-1 + 2=1).

Step4: Check the continuity condition

A function (y = f(x)) is continuous at (x=a) if (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)). Here, (\lim_{x\rightarrow1^{-}}f(x)=1), (\lim_{x\rightarrow1^{+}}f(x)=3), and since (1\neq3), the function is not continuous.