Question

for what value of c is the function defined below continuous on $(-\infty,\infty)$?
$f(x)=\

$$\begin{cases}x^{2}-c^{2},&x < 4,\\crx + 20,&x\\geq4.\\end{cases}$$

$

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = 4$, $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{-}}(x^{2}-c^{2})=4^{2}-c^{2}=16 - c^{2}$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow4^{+}}f(x)=\lim_{x
ightarrow4^{+}}(cx + 20)=4c+20$.

Step4: Set limits equal

Set $16 - c^{2}=4c + 20$. Rearrange to get $c^{2}+4c + 4 = 0$.

Step5: Solve quadratic equation

Factor $c^{2}+4c + 4=(c + 2)^{2}=0$. So $c=-2$.

Answer:

$-2$