ACT Math Quiz: Piecewise Functions
20 questions · exam conditions
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Piecewise FunctionsQuestion 1 of 20

Which interval contains x=−1x = -1 for the piecewise function f(x)={3x+7if x≤−1x2−2if x>−1f(x) = \begin{cases} 3x + 7 & \text{if } x \leq -1 \\ x^2 - 2 & \text{if } x > -1 \end{cases}?

x<−1x < -1
x≤−1x \leq -1
x>−1x > -1
x=0x = 0
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ACT Math Quiz

ACT Math Quiz: Piecewise Functions

Practice Piecewise Functions in ACT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Piecewise Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which interval contains x=−1x = -1 for the piecewise function f(x)={3x+7if x≤−1x2−2if x>−1f(x) = \begin{cases} 3x + 7 & \text{if } x \leq -1 \\ x^2 - 2 & \text{if } x > -1 \end{cases}?

  1. x<−1x < -1
  2. x≤−1x \leq -1 (correct answer)
  3. x>−1x > -1
  4. x=0x = 0
Explanation: To determine which interval contains x=−1x = -1, we check each condition: Is −1≤−1-1 \leq -1? Yes. Is −1>−1-1 > -1? No. Since −1-1 satisfies the condition x≤−1x \leq -1, it belongs to the first interval. The boundary point x=−1x = -1 is included in the first piece due to the ≤\leq symbol.

Question 2

What is f(0)f(0) for the piecewise function f(x)={2x+4if x<1x2−6if x≥1f(x) = \begin{cases} 2x + 4 & \text{if } x < 1 \\ x^2 - 6 & \text{if } x \geq 1 \end{cases}?

  1. 4 (correct answer)
  2. 2
  3. 6
  4. 0
Explanation: For x = 0, we check the intervals: Is 0 < 1? Yes. So we use the first piece: f(x)=2x+4f(x) = 2x + 4. Substituting x = 0: f(0)=2(0)+4=0+4=4f(0) = 2(0) + 4 = 0 + 4 = 4. Choice B would result from using the second piece incorrectly.

Question 3

A savings plan applies a rule f(x)f(x) to the number of weeks xx you have saved. For the piecewise function f(x)={6−xif x<42x+1if 4≤x<9x2−10if x≥9f(x)=\begin{cases}6-x & \text{if } x<4\\ 2x+1 & \text{if } 4\le x<9\\ x^2-10 & \text{if } x\ge 9\end{cases} what is f(9)f(9)?

  1. 19
  2. 81
  3. 8
  4. 71 (correct answer)
Explanation: For x = 9, determine which piece to use: Is 9 < 4? No. Is 4 ≤ 9 < 9? No, since 9 is not less than 9. Is 9 ≥ 9? Yes. Use the third piece: f(x) = x² - 10. Thus f(9) = 9² - 10 = 81 - 10 = 71.

Question 4

Which interval contains x = 3 for the function $$f(x) = \begin{cases} 3x + 1 & \text{if } x < 1 \ 2x - 2 & \text{if } 1 \leq x < 4 \ x^2 & \text{if } x \geq 4 \end{cases}

  1. x<1x < 1
  2. 1≤x<41 \leq x < 4 (correct answer)
  3. x≥4x \geq 4
  4. x>4x > 4
Explanation: For x=3x = 3, check each interval: 3<13 < 1? No. 1≤3<41 \leq 3 < 4? Yes, since 1≤31 \leq 3 and 3<43 < 4. 3≥43 \geq 4? No. Therefore, x=3x = 3 falls in the interval 1≤x<41 \leq x < 4.

Question 5

What is f(2) for the piecewise function: $$f(x) = \begin{cases} -x + 3 & \text{if } x < 1 \ 4x & \text{if } 1 \leq x < 3 \ x^2 - 1 & \text{if } x \geq 3 \end{cases}

  1. 8 (correct answer)
  2. 7
  3. 9
  4. 6
Explanation: For x = 2, check intervals: 2 < 1? No. 1 ≤ 2 < 3? Yes. So use the second piece f(x)=4xf(x) = 4x. Substitute x = 2: f(2)=4(2)=8f(2) = 4(2) = 8. The value x = 2 falls clearly within the middle interval.

Question 6

A company assigns a performance rating f(x)f(x) based on an employee's score xx. The rating function is

7-x & \text{if } x<0 \\ 3x+1 & \text{if } 0\le x<4 \\ 15 & \text{if } x\ge 4 \end{cases}

Based on the piecewise function, what is the value when x=0x=0?

  1. 7
  2. 1 (correct answer)
  3. 0
  4. 15
Explanation: For x = 0, check intervals: Is 0 < 0? No. Is 0 ≤ 0 < 4? Yes, since 0 = 0 satisfies this condition. Use the second piece: f(x) = 3x + 1. Substituting: f(0) = 3(0) + 1 = 0 + 1 = 1. The boundary x = 0 falls in the middle piece due to the ≤ sign.

Question 7

A game assigns points f(x)f(x) based on a player's level xx using the piecewise function below. For

3x+2 & \text{if } x<2\\ 10 & \text{if } 2\le x<5\\ -x+20 & \text{if } x\ge 5 \end{cases}

what is f(2)f(2)?

  1. 8
  2. 10 (correct answer)
  3. 12
  4. 18
Explanation: For x = 2, check which interval contains 2: Is 2 < 2? No. Is 2 ≤ 2 < 5? Yes, since 2 ≤ 2 is true and 2 < 5 is true. Therefore, use the second piece f(x) = 10. Since this piece is a constant function, f(2) = 10. The boundary x = 2 belongs to the middle interval due to the ≤ sign.

Question 8

A machine's output f(x)f(x) depends on the setting xx using the piecewise function below. For

-x+6 & \text{if } x<1\\ 2x & \text{if } 1\le x<6\\ x^2-10 & \text{if } x\ge 6 \end{cases}

what is f(0)f(0)?

  1. 0
  2. 6 (correct answer)
  3. 12
  4. -6
Explanation: For x = 0, check intervals: Is 0 < 1? Yes. Therefore, use the first piece f(x) = -x + 6. Substituting x = 0: f(0) = -0 + 6 = 6. Since 0 is less than 1, we don't need to check the other intervals.

Question 9

A grading policy assigns a score adjustment f(x)f(x) based on the raw score xx. For the piecewise function

-x & \text{if } x<-1\\ 2x+5 & \text{if } -1\le x<3\\ 11 & \text{if } x\ge 3 \end{cases}

what is f(−1)f(-1)?

  1. 3 (correct answer)
  2. 1
  3. 7
  4. 11
Explanation: For x = -1, check intervals: Is -1 < -1? No. Is -1 ≤ -1 < 3? Yes, since -1 ≤ -1 is true and -1 < 3 is true. So use the second piece f(x) = 2x + 5. Substituting x = -1: f(-1) = 2(-1) + 5 = -2 + 5 = 3. The boundary x = -1 belongs to the middle interval due to the ≤ sign.

Question 10

Which interval contains x=−2x = -2 for the piecewise function f(x)={x2if x≤−23x+5if x>−2f(x) = \begin{cases} x^2 & \text{if } x \leq -2 \\ 3x + 5 & \text{if } x > -2 \end{cases}?

  1. x≥−1x \geq -1
  2. x>−2x > -2
  3. x<−3x < -3
  4. x≤−2x \leq -2 (correct answer)
Explanation: To determine which interval contains x=−2x = -2, we check each condition: Is −2≤−2-2 \leq -2? Yes. Is −2>−2-2 > -2? No. Since −2-2 satisfies the condition x≤−2x \leq -2, it belongs to the first interval. The boundary point x=−2x = -2 is included in the first piece due to the ≤\leq symbol.

Question 11

For the function f(x)={2x−5if x<−2x2+3if x≥−2f(x) = \begin{cases} 2x - 5 & \text{if } x < -2 \\ x^2 + 3 & \text{if } x \geq -2 \end{cases}, what is f(−3)f(-3)?

  1. -11 (correct answer)
  2. -1
  3. -9
  4. -7
Explanation: For x = -3, we check which interval applies: Is -3 < -2? Yes. So we use the first piece: f(x)=2x−5f(x) = 2x - 5. Substituting x = -3: f(−3)=2(−3)−5=−6−5=−11f(-3) = 2(-3) - 5 = -6 - 5 = -11. Choice D would result from using the second piece incorrectly.

Question 12

For the piecewise function f(x)={x2+3if x≤02x−4if x>0f(x) = \begin{cases} x^2 + 3 & \text{if } x \leq 0 \\ 2x - 4 & \text{if } x > 0 \end{cases}, what is f(1)f(1)?

  1. -2 (correct answer)
  2. 5
  3. -1
  4. 4
Explanation: For x = 1, we check which interval applies: Is 1≤01 \leq 0? No. Is 1>01 > 0? Yes. So we use the second piece: f(x)=2x−4f(x) = 2x - 4. Substituting x = 1: f(1)=2(1)−4=2−4=−2f(1) = 2(1) - 4 = 2 - 4 = -2. Choice B would result from using the first piece incorrectly.

Question 13

For the function $$f(x) = \begin{cases} 2 - x & \text{if } x < 0 \ 3x + 1 & \text{if } 0 \leq x < 3 \ x^2 & \text{if } x \geq 3 \end{cases}

  1. 12
  2. 10
  3. 9 (correct answer)
  4. 11
Explanation: For x=3x = 3, check intervals: 3<03 < 0? No. 0≤3<30 \leq 3 < 3? No. 3≥33 \geq 3? Yes. So use the third piece f(x)=x2f(x) = x^2. Substitute x=3x = 3: f(3)=32=9f(3) = 3^2 = 9. Note that x=3x = 3 falls in the third piece due to the ≥\geq condition.

Question 14

A game assigns points based on your score xx using the piecewise function shown. For the function f(x)={x−7if x<03x+2if 0≤x<4x2if x≥4f(x)=\begin{cases}x-7 & \text{if } x<0\\ 3x+2 & \text{if } 0\le x<4\\ x^2 & \text{if } x\ge 4\end{cases} what is f(4)f(4)?

  1. 14
  2. 12
  3. 16 (correct answer)
  4. -3
Explanation: For x = 4, check intervals: Is 4 < 0? No. Is 0 ≤ 4 < 4? No, since 4 is not less than 4. Is 4 ≥ 4? Yes. So we use the third piece: f(x) = x². Thus f(4) = 4² = 16. Note the boundary: x = 4 falls in the third piece due to the ≥ condition.

Question 15

A taxi company models a surcharge f(x)f(x) based on time xx (in minutes) with the piecewise function below. For

4 & \text{if } x<2\\ 3x+1 & \text{if } 2\le x<8\\ 25-x & \text{if } x\ge 8 \end{cases}

what is f(8)f(8)?

  1. 25
  2. 17 (correct answer)
  3. 18
  4. 4
Explanation: For x = 8, check which interval contains 8: Is 8 < 2? No. Is 2 ≤ 8 < 8? No, since 8 < 8 is false. Is 8 ≥ 8? Yes. Therefore, use the third piece f(x) = 25 - x. Substituting x = 8: f(8) = 25 - 8 = 17. The boundary x = 8 belongs to the third interval due to the ≥ sign.

Question 16

What is f(-3) for the piecewise function: $$f(x) = \begin{cases} 2x^2 & \text{if } x < 0 \ x + 4 & \text{if } 0 \leq x < 3 \ 3x - 2 & \text{if } x \geq 3 \end{cases}

  1. 18 (correct answer)
  2. 7
  3. 2
  4. 20
Explanation: For x = -3, check intervals: -3 < 0? Yes. So use the first piece f(x) = 2x22x^2. Substitute x = -3: f(−3)=2(−3)2=2(9)=18f(-3) = 2(-3)^2 = 2(9) = 18. Since -3 is negative, it clearly falls in the first interval.

Question 17

For the function f(x)={3x−7if x≤1x2−3if x>1f(x) = \begin{cases} 3x - 7 & \text{if } x \leq 1 \\ x^2 - 3 & \text{if } x > 1 \end{cases}, what is f(1)f(1)?

  1. -4 (correct answer)
  2. -3
  3. -2
  4. 0
Explanation: For x = 1, we check the intervals: Is 1 ≤ 1? Yes. So we use the first piece: f(x)=3x−7f(x) = 3x - 7. Substituting x = 1: f(1)=3(1)−7=3−7=−4f(1) = 3(1) - 7 = 3 - 7 = -4. The boundary point x = 1 is included in the first piece due to the ≤ symbol.

Question 18

What is f(−2)f(-2) for the piecewise function f(x)={2x+3if x<0−x2+2if x≥0f(x) = \begin{cases} 2x + 3 & \text{if } x < 0 \\ -x^2 + 2 & \text{if } x \geq 0 \end{cases}?

  1. -1 (correct answer)
  2. -4
  3. -2
  4. -3
Explanation: For x = -2, we check which interval applies: Is -2 < 0? Yes. So we use the first piece: f(x) = 2x + 3. Substituting x = -2: f(-2) = 2(-2) + 3 = -4 + 3 = -1. Choice B would result from using the second piece incorrectly.

Question 19

For the piecewise function

3x+2, & x<2\\ 8-x, & 2\le x<6\\ (x-6)^2, & x\ge 6 \end{cases}

what is f(5)f(5)?

  1. 17
  2. 3 (correct answer)
  3. 25
  4. 13
Explanation: For x=5x = 5, check which interval contains this value: 5<25 < 2? No. 2≤5<62 \leq 5 < 6? Yes. Since x=5x = 5 satisfies the condition 2≤x<62 \leq x < 6, we use the second piece f(x)=8−xf(x) = 8 - x. Substituting x=5x = 5: f(5)=8−5=3f(5) = 8 - 5 = 3. Choice A might result from incorrectly using the first piece or misreading the intervals.

Question 20

What is f(−3)f(-3) for the function defined as f(x)={5x+1if x<−23x2−xif x≥−2f(x) = \begin{cases} 5x + 1 & \text{if } x < -2 \\ 3x^2 - x & \text{if } x \geq -2 \end{cases}?

  1. -14 (correct answer)
  2. -8
  3. -16
  4. -12
Explanation: For x = -3, we check the intervals: Is -3 < -2? Yes. So we use the first piece: f(x)=5x+1f(x) = 5x + 1. Substituting x = -3: f(−3)=5(−3)+1=−15+1=−14f(-3) = 5(-3) + 1 = -15 + 1 = -14. Choice D would result from using the second piece incorrectly.