Question

work-out problem 2, 10 points
use the three - part definition of continuity to determine whether the given function is continuous at $x = 4$.
$f(x)=\begin{cases}x + 8&\text{if }x<4\\2x^{2}-16&\text{if }xgeq4end{cases}$

Explanation:

Step1: Check if $f(4)$ exists

Substitute $x = 4$ into the second - part of the piece - wise function since $x=4$ satisfies $x\geq4$. So, $f(4)=2(4)^{2}-16=2\times16 - 16=32 - 16 = 16$.

Step2: Calculate $\lim_{x

ightarrow4^{-}}f(x)$
As $x
ightarrow4^{-}$, we use the first - part of the piece - wise function $f(x)=x + 8$. Then $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{-}}(x + 8)=4 + 8=12$.

Step3: Calculate $\lim_{x

ightarrow4^{+}}f(x)$
As $x
ightarrow4^{+}$, we use the second - part of the piece - wise function $f(x)=2x^{2}-16$. Then $\lim_{x
ightarrow4^{+}}f(x)=\lim_{x
ightarrow4^{+}}(2x^{2}-16)=2(4)^{2}-16=16$.

Step4: Check the continuity condition

For a function to be continuous at $x = a$, we need $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here, $\lim_{x
ightarrow4^{-}}f(x)=12$, $\lim_{x
ightarrow4^{+}}f(x)=16$ and $f(4)=16$. Since $\lim_{x
ightarrow4^{-}}f(x)
eq\lim_{x
ightarrow4^{+}}f(x)$, the function is not continuous at $x = 4$.

Answer:

The function $f(x)$ is not continuous at $x = 4$.