Answer
(a) At $x=2$, the graph is continuous.
At $x=3$, the graph is continuous.
At $x=4$, the graph is continuous from the right.
(b) We can see a sketch of the graph below.
Work Step by Step
(a) $g(2) = 2(2)- (2)^2 = 0$
$g(3) = 2-(3) = -1$
$g(4) = \pi$
$\lim\limits_{x \to 2^-}g(x) = \lim\limits_{x \to 2^-}[2(x) - x^2] = 0$
$\lim\limits_{x \to 2^+}g(x) = \lim\limits_{x \to 2^+}(2-x) = 0$
$\lim\limits_{x \to 3^-}g(x) = \lim\limits_{x \to 3^-}(2-x) = -1$
$\lim\limits_{x \to 3^+}g(x) = \lim\limits_{x \to 3^+}(x-4) = -1$
$\lim\limits_{x \to 4^-}g(x) = \lim\limits_{x \to 4^-}(x-4) = 0$
$\lim\limits_{x \to 4^+}g(x) = \lim\limits_{x \to 4^+}\pi = \pi$
At $x=2$, the graph is continuous
$\lim\limits_{x \to 2}g(x) = g(2)$
At $x=3$, the graph is continuous.
$\lim\limits_{x \to 3}g(x) = g(3)$
At $x=4$, the graph is continuous from the right.
$\lim\limits_{x \to 4^+}g(x) = g(4)$
(b) We can see a sketch of the graph below.
