Question

question 17 - 1 point
find the points at which the following function is discontinuous.
$f(x)=\

$$\begin{cases}-5x & \\text{if }x < - 5 \\\\2x^{2}-25 & \\text{if }-5 < x < - 3 \\\\6x^{2}-61 & \\text{if }x\\geq - 3\\end{cases}$$

$
select the correct answer below:
o $f(x)$ is discontinuous at - 5.
o $f(x)$ is discontinuous at - 3.
o none of the above
feedback

Explanation:

Step1: Recall continuity condition

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

Step2: Check continuity at $x=-5$

For $x
ightarrow - 5^{-}$, $f(x)=-5x$. So, $\lim_{x
ightarrow - 5^{-}}f(x)=-5\times(-5) = 25$.
For $x
ightarrow - 5^{+}$, $f(x)=2x^{2}-25$. So, $\lim_{x
ightarrow - 5^{+}}f(x)=2\times(-5)^{2}-25=2\times25 - 25=25$. And $f(-5)=2\times(-5)^{2}-25 = 25$. So, $f(x)$ is continuous at $x = - 5$.

Step3: Check continuity at $x=-3$

For $x
ightarrow - 3^{-}$, $f(x)=2x^{2}-25$. So, $\lim_{x
ightarrow - 3^{-}}f(x)=2\times(-3)^{2}-25=2\times9 - 25=-7$.
For $x
ightarrow - 3^{+}$, $f(x)=6x^{2}-61$. So, $\lim_{x
ightarrow - 3^{+}}f(x)=6\times(-3)^{2}-61=6\times9 - 61=-7$. And $f(-3)=6\times(-3)^{2}-61=-7$. So, $f(x)$ is continuous at $x=-3$.

Answer:

None of the above