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

Explanation:

Step1: Find the value of the function at $x = 1$

For $x\leq1$, $f(1)=-1 + 2=1$.

Step2: Find the left - hand limit as $x\to1$

$\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{-}}(-x + 2)=-1+2 = 1$.

Step3: Find the right - hand limit as $x\to1$

For $x>1$, $\lim_{x\to1^{+}}f(x)=\lim_{x\to1^{+}}(5x - 2)=5\times1-2=3$.

Step4: Check the continuity condition

Since $\lim_{x\to1^{-}}f(x)=1$, $\lim_{x\to1^{+}}f(x)=3$ and $f(1) = 1$, and $\lim_{x\to1^{-}}f(x)\neq\lim_{x\to1^{+}}f(x)$, the function is not continuous at $x = 1$.

Answer:

The function $f(x)$ is not continuous.