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Question
suppose that the function $f$ is defined, for all real numbers, as follows.
$f(x)=\begin{cases}-x + 2&\text{if }xleq1\\5x - 2&\text{if }x>1end{cases}$
graph the function $f$. then determine whether or not the function is continuous.
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 $\lim_{x\to1^{-}}f(x)
eq\lim_{x\to1^{+}}f(x)$, the function is not continuous at $x = 1$.
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The function $f(x)$ is not continuous.