Question

suppose that the function f is defined for all real numbers as follows.
f(x)=\begin{cases}-10 + x^{2}&\text{if }-4leq x<4\\-x + 10&\text{if }xgeq4end{cases}
graph the function f. then determine whether or not the function is continuous.

Explanation:

Step1: Analyze $f(x)=- 10 + x^{2}$ for $-4\leq x<4$

This is a parabola. When $x = - 4$, $y=-10+(-4)^{2}=-10 + 16=6$. When $x = 4$, $y=-10 + 4^{2}=-10+16 = 6$. The vertex of the parabola $y=-10 + x^{2}$ is at $(0,-10)$.

Step2: Analyze $f(x)=-x + 10$ for $x\geq4$

When $x = 4$, $y=-4 + 10=6$. As $x$ increases, $y$ decreases. For example, when $x=10$, $y=-10 + 10=0$.

Step3: Check continuity at $x = 4$

Left - hand limit as $x\to4^{-}$: $\lim_{x\to4^{-}}(-10 + x^{2})=-10+16 = 6$. Right - hand limit as $x\to4^{+}$: $\lim_{x\to4^{+}}(-x + 10)=-4 + 10=6$. And $f(4)=-4 + 10=6$. Since $\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{+}}f(x)=f(4)$, the function is continuous.

Answer:

The function is continuous.