Question

Locate the points of discontinuity in the piecewise function shown below:

|-(x + 1)² + 2; -[infinity] < x < -1
f(x) = {-x + 2; -1 ≤ x < 2
√x - 1; 2 ≤ x < [infinity]

a) x = -1 and 2
b) x = 2
c) x = -1
d) No points of discontinuity

Answer 1

The piecewise function given has one point of discontinuity at x = 2, as the value of the function changes between the adjacent formulas at this point.

Step-by-step explanation:

To locate the points of discontinuity in the given piecewise function, we examine the function's behavior at the points where the formula changes. These points are at x = -1 and x = 2. For the function f(x) to be continuous at these points, the left-hand limit and the right-hand limit must approach the same value as the point in question, and the function itself must be defined at that point.

For x = -1, we need to compare the limit of -(x + 1)² + 2 as x approaches -1 from the left with the value of -x + 2 at x = -1. For x = 2, we must compare the limit of -x + 2 as x approaches 2 from the left with the value of √x - 1 at x = 2.

Upon examination, we find that at x = -1, -(x + 1)² + 2 equals 2, and -x + 2 also equals 2, meaning the function is continuous at x = -1. At x = 2, -x + 2 equals 0 while √x - 1 equals 1, indicating a discontinuity since the values do not match up at x = 2. Therefore, the only point of discontinuity is x = 2.