1) Let f(x) = -2x + 1 : x > 2, -2x^2 + 5 : -1 < x ≤ 2, -4x - 5 : x ≤ -1.

Question

(1) Let

f(x) = -2x + 1 x > 2

-2x^2 + 5 -1 < x ≤ 2

-4x - 5 x ≤ -1.

(a) Graph f(x).

(b) State all points where f(x) is discontinuous and why.

(c) For which of the points found in part (a), is the function right continuous?

(d) For which of the points in part (a), is the function left continuous?

Answer 1

First draw the coordinate plane and the two boundaries, x = -1 and x = 2. Then draw the first graph , y = -2x + 1 and erase the part to the left of x = 2. Like this:

Then draw the second function, y = -2x² + 5 and erase the part to the left of x = -1 and to the right of x = 2. Like this:

Notice that the two sections of graph overlap at the point (2, -3) so this dot would now be colored in (not because of y = -2x + 1 because that portion does not exist at x = 2, but because of y = -2x² + 5).

Then draw the last section of graph, y = -4x - 5 and erase the part to the right of x = -1.

Looking at the graph, you can see that it is not continuous at x = -1. You should be able to answer the other questions by just looking at the graph.

Answer 2