Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 24

Suppose that

$\qquad f(x) = \left\{ \begin{array}{ccc} 2x - x^2 & \text{ if } & 0 \leq x \leq 2 \\ 2 - x & \text{ if } & 2 < x \leq 3 \\ x - 4 & \text{ if } & 3 < x < 4 \\ \pi & \text{ if } & x \geq 4 \end{array} \right.$

a.) Determine whether $f$ is continuous from the left, continuous from the right, or continuous at each of the numbers 2, 3 and 4.

$\qquad$ We evaluate the left and right hand limits for the numbers 2, 3, and 4 to see whether $f$ is continuoous on that point.

$\qquad$ at $x = 2$,

$\qquad \qquad$ left hand limit: $\lim \limits_{x \to 2^-} f(x) = \lim \limits_{x \to 2^-} 2x - x^2 = 2(2) - 2(2)^2 = 4-4 = 0$

$\qquad \qquad$ right hand limit: $\lim \limits_{x \to 2^+} f(x) = \lim \limits_{x \to 2^+} 2 - x = 2 - 2 = 0$

$\qquad \qquad$ Therefore, $f$ is continuous at number 2.

$\qquad$ at $x = 3,$

$\qquad \qquad$ left hand limit: $\lim \limits_{x \to 3^-} f(x) = \lim \limits_{x \to 3^-} 2 -x = 2 - (3) = -1$

$\qquad \qquad$ right hand limit: $\lim \limits_{x \to 3^+} f(x) = \lim \limits_{x \to 3^+} x - 4 = 3 - 4 = -1$

$\qquad \qquad$ Therefore, $f$ is continuous at number 3.

$\qquad$ at $x = 4$,

$\qquad \qquad$ left hand limit: $\lim \limits_{x \to 4^-} f(x) = \lim \limits_{x \to 4^-} x- 4 = 4 - 4 =0$

$\qquad \qquad$ right hand limit: $\lim \limits_{x \to 4^+} f(x) = \lim \limits_{x \to 4^+} \pi = \pi$

$\qquad \qquad$ The left and right hand limit of the function at number 4 are different. Therefore, $f$ is discontinuous at 4 but it is continuous from the left.

b.) Graph the function $f$.