Question

example 3: given the function, $f(x)=\begin{cases}\frac{1}{2}x + 1; &if x < 4\\-x + 7; &if xgeq4end{cases}$ a. is $f(x)$ continuous at 4, why or why not?

Explanation:

Step1: Find the left - hand limit

We find $\lim_{x
ightarrow4^{-}}f(x)$. Since $x
ightarrow4^{-}$ (approaching 4 from the left), we use $f(x)=\frac{1}{2}x + 1$.
$\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{-}}(\frac{1}{2}x + 1)=\frac{1}{2}\times4+1=2 + 1=3$.

Step2: Find the right - hand limit

We find $\lim_{x
ightarrow4^{+}}f(x)$. Since $x
ightarrow4^{+}$ (approaching 4 from the right), we use $f(x)=-x + 7$.
$\lim_{x
ightarrow4^{+}}f(x)=\lim_{x
ightarrow4^{+}}(-x + 7)=-4 + 7=3$.

Step3: Find the function value at $x = 4$

We find $f(4)$. Since $x = 4$, we use $f(x)=-x + 7$. So $f(4)=-4 + 7=3$.

Step4: Check the continuity condition

A function $y = f(x)$ is continuous at $x=a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here, $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{+}}f(x)=f(4)=3$.

Answer:

Yes, $f(x)$ is continuous at $x = 4$ because $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{+}}f(x)=f(4)=3$.