Question

work-out problem 3, 10 points
sketch a graph of a function g for which:
g(-2)=0, g(-1)=4, g(0)=3, g(1)=-2, g(2)=0, g(3)=2
lim_{x→ - 3^{+}}g(x)=-∞, lim_{x→4^{-}}f(x)=∞, lim_{x→1}g(x)=2, lim_{x→3^{-}}g(x)=-4, lim_{x→3^{+}}g(x)=2
g(-2)=4, g(-1)=0, g(2)=-4
5 - point bonus
give the two x - values where g is discontinuous.

Explanation:

Step1: Recall continuity condition

A function $g(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a}g(x)=g(a)$.

Step2: Analyze limits and function - values

We know that $\lim_{x
ightarrow - 3^{+}}g(x)=-\infty$. Since the limit is not a finite value, the function is discontinuous at $x=-3$. Also, $\lim_{x
ightarrow4^{-}}f(x)=\infty$ (it should be $\lim_{x
ightarrow4^{-}}g(x)=\infty$ probably, assuming it's a typo). Since the limit as $x$ approaches $4$ from the left is not a finite value, the function is discontinuous at $x = 4$.

Answer:

$x=-3,x = 4$