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ACT Math Help: Piecewise Functions

Review real example questions for Piecewise Functions in ACT Math.

Question 1 / 10

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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}?

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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.