Question

for what value of the constant c is the function f continuous on the interval (-∞,∞).
f(x)=\begin{cases}x^{2}-9, & x < c\\8x - 25, & xgeq cend{cases}

Explanation:

Step1: Recall continuity condition

For a function $f(x)$ to be continuous at $x = c$, $\lim_{x
ightarrow c^{-}}f(x)=\lim_{x
ightarrow c^{+}}f(x)=f(c)$. Here, $\lim_{x
ightarrow c^{-}}f(x)=c^{2}-9$ and $\lim_{x
ightarrow c^{+}}f(x)=8c - 25$.

Step2: Set left - hand and right - hand limits equal

Set $c^{2}-9=8c - 25$. Rearrange to get a quadratic equation: $c^{2}-8c + 16 = 0$.

Step3: Solve the quadratic equation

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For $c^{2}-8c + 16 = 0$, where $a = 1$, $b=-8$, $c = 16$, we have $c=\frac{8\pm\sqrt{(-8)^{2}-4\times1\times16}}{2\times1}=\frac{8\pm\sqrt{64 - 64}}{2}=\frac{8\pm0}{2}$. So $c = 4$.

Answer:

$4$