Piecewise Functions Worksheet - Free Printable

1. For the algebraic piecewise function problem, the graph shows three segments:

- For x < -1: The line passes through (-2, -1) and (-1, 1), so the slope is (1 - (-1))/(-1 - (-2)) = 2/1 = 2. Using point-slope form with (-2, -1): y - (-1) = 2(x - (-2)) → y + 1 = 2x + 4 → y = 2x + 3. So the equation is y = 2x + 3 for x < -1.
- For -1 ≤ x ≤ 1: The graph is a horizontal line at y = 1. So the equation is y = 1 for -1 ≤ x ≤ 1.
- For x > 1: The line passes through (1, 1) and (3, 3), so the slope is (3 - 1)/(3 - 1) = 2/2 = 1. Using point-slope form with (1, 1): y - 1 = 1(x - 1) → y = x. So the equation is y = x for x > 1.

The piecewise function is:
f(x) = { 2x + 3, if x < -1
{ 1, if -1 ≤ x ≤ 1
{ x, if x > 1

The domain for each piece is:
- For y = 2x + 3: domain is (-∞, -1)
- For y = 1: domain is [-1, 1]
- For y = x: domain is (1, ∞)

2. For the verbal problem about t-shirt costs:

The piecewise function for the cost per shirt f(x) where x is the number of shirts is:
f(x) = { 25, if 0 ≤ x ≤ 20
{ 20, if 21 ≤ x ≤ 30
{ 15, if 31 ≤ x ≤ 50
{ 10, if x ≥ 51

To graph this:
- Draw a horizontal line at y = 25 from x = 0 to x = 20 (inclusive). Use closed circles at (0, 25) and (20, 25).
- Draw a horizontal line at y = 20 from x = 21 to x = 30 (inclusive). Use closed circles at (21, 20) and (30, 20).
- Draw a horizontal line at y = 15 from x = 31 to x = 50 (inclusive). Use closed circles at (31, 15) and (50, 15).
- Draw a horizontal line at y = 10 for x ≥ 51. Use a closed circle at (51, 10) and extend the line to the right.

3. For the SAT prep multiple choice question:
The regulation requires a child's height h to be between 30 inches and 50 inches, so 30 ≤ h ≤ 50.
We need to find which inequality represents this.
- (A) |h - 10| < 50 → -50 < h - 10 < 50 → -40 < h < 60. This is too broad.
- (B) |h - 20| < 40 → -40 < h - 20 < 40 → -20 < h < 60. Too broad.
- (C) |h - 30| < 20 → -20 < h - 30 < 20 → 10 < h < 50. This excludes h = 30 and includes h > 50, so incorrect.
- (D) |h - 40| < 10 → -10 < h - 40 < 10 → 30 < h < 50. This excludes h = 30 and h = 50, so incorrect.
- (E) |h - 45| < 5 → -5 < h - 45 < 5 → 40 < h < 50. Too narrow.

Wait, none of these seem to include both 30 and 50. But let's check again. The correct inequality should be |h - 40| < 10? No, that's 30 < h < 50. But the regulation is 30 ≤ h ≤ 50. So perhaps we need to check the options again.

Actually, the correct inequality should be |h - 40| ≤ 10 → 30 ≤ h ≤ 50. But that's not an option. Let's recheck the options.

Looking at (E) |h - 45| < 5 → 40 < h < 50. Not correct.

Wait, perhaps I made a mistake. Let's check (D) |h - 40| < 10 → 30 < h < 50. This is almost correct but excludes the endpoints. But the regulation is "between 30 inches and 50 inches tall", which might include the endpoints. However, in multiple choice, we need to see which one is closest.

Actually, let's check the midpoint. The interval [30, 50] has midpoint 40, and length 20, so the distance from the midpoint is less than 10. So |h - 40| < 10 → 30 < h < 50. But the regulation says "between 30 and 50", which could be interpreted as exclusive. But typically, such regulations include the endpoints. However, since none of the options include the endpoints, perhaps the intended answer is (D) |h - 40| < 10, assuming "between" means exclusive.

But let's check the options again. Option (D) is |h - 40| < 10 → 30 < h < 50. Option (E) is |h - 45| < 5 → 40 < h < 50. So (D) is closer.

Wait, the correct answer should be (D) |h - 40| < 10, as it represents the interval (30, 50), which is the closest to [30, 50]. But the regulation says "between 30 inches and 50 inches tall", which usually includes the endpoints. But since the options don't have ≤, and (D) is the only one that covers the interval correctly if we assume "between" is exclusive, perhaps that's the intended answer.

Actually, let's check the midpoint of 30 and 50 is 40, and the distance from 40 to 30 or 50 is 10, so |h - 40| < 10 gives 30 < h < 50. So (D) is correct if the regulation is exclusive. But the problem says "between 30 inches and 50 inches tall", which in common usage includes the endpoints. However, in mathematics, "between a and b" can be ambiguous. But since the options don't have ≤, and (D) is the only one that covers the interval (30, 50), and the others are incorrect, (D) is the best answer.

But let's check the options again. Option (C) |h - 30| < 20 → 10 < h < 50. This includes heights less than 30, so incorrect.

Option (E) |h - 45| < 5 → 40 < h < 50. This is too narrow.

So the correct answer is (D) |h - 40| < 10.

4. For the grid-in problem:
Given:
2|x - 9| = 24
|xy| = 15
x < 0 < y

From 2|x - 9| = 24 → |x - 9| = 12
So x - 9 = 12 or x - 9 = -12
→ x = 21 or x = -3

But x < 0, so x = -3

Then |xy| = 15 → |-3 * y| = 15 → 3|y| = 15 → |y| = 5
Since y > 0, y = 5

So x + y = -3 + 5 = 2

The answer is 2.