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ACT Math · Learn by Concept

ACT Math Help: Piecewise Functions

Review real example questions for Piecewise Functions in ACT Math.

Question 1 / 10

0 of 10 answered

Which interval contains x=−1x = -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}f(x)={3x+7x2−2​if x≤−1if x>−1​?

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All questions

Question 1

Which interval contains x=−1x = -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}f(x)={3x+7x2−2​if x≤−1if x>−1​?

  1. x<−1x < -1x<−1
  2. x≤−1x \leq -1x≤−1 (correct answer)
  3. x>−1x > -1x>−1
  4. x=0x = 0x=0

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

Question 2

What is f(0)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}f(x)={2x+4x2−6​if x<1if x≥1​?

  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 + 4f(x)=2x+4. Substituting x = 0: f(0)=2(0)+4=0+4=4f(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)f(x) to the number of weeks xxx 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}f(x)=⎩⎨⎧​6−x2x+1x2−10​if x<4if 4≤x<9if x≥9​ what is f(9)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

For the piecewise function f(x)={x+3if x<−2−2xif −2≤x<3x2−5if x≥3f(x) = \begin{cases} x + 3 & \text{if } x < -2 \\ -2x & \text{if } -2 \leq x < 3 \\ x^2 - 5 & \text{if } x \geq 3 \end{cases}f(x)=⎩⎨⎧​x+3−2xx2−5​if x<−2if −2≤x<3if x≥3​, what is f(3)?

  1. 9
  2. 4 (correct answer)
  3. 5
  4. 3

Explanation: For x=3x = 3x=3, check intervals: 3<−23 < -23<−2? No. −2≤3<3-2 \leq 3 < 3−2≤3<3? No. 3≥33 \geq 33≥3? Yes. So use the third piece f(x)=x2−5f(x) = x^2 - 5f(x)=x2−5. Substitute x=3x = 3x=3: f(3)=32−5=9−5=4f(3) = 3^2 - 5 = 9 - 5 = 4f(3)=32−5=9−5=4. Note that x=3x = 3x=3 falls in the third piece due to the ≥\geq≥ condition.

Question 5

What is f(1) for the piecewise function: f(x)={x2−1if x<13if 1≤x<44xif x≥4f(x) = \begin{cases} x^2 - 1 & \text{if } x < 1 \\ 3 & \text{if } 1 \leq x < 4 \\ 4x & \text{if } x \geq 4 \end{cases}f(x)=⎩⎨⎧​x2−134x​if x<1if 1≤x<4if x≥4​?

  1. 1
  2. 0
  3. 3 (correct answer)
  4. 4

Explanation: For x = 1, check intervals: 1 < 1? No. 1 ≤ 1 < 4? Yes. So use the second piece f(x)=3f(x) = 3f(x)=3. At the boundary x = 1, we use the second piece due to the ≤ condition. Therefore f(1)=3f(1) = 3f(1)=3.

Question 6

Which interval contains x = 3 for the function f(x)={3x+1if x<12x−2if 1≤x<4x2if x≥4f(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}f(x)=⎩⎨⎧​3x+12x−2x2​if x<1if 1≤x<4if x≥4​?

  1. x<1x < 1x<1
  2. 1≤x<41 \leq x < 41≤x<4 (correct answer)
  3. x≥4x \geq 4x≥4
  4. x>4x > 4x>4

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

Question 7

Based on the piecewise function f(x)={2x−3if x<−1x2+2if −1≤x<25x−1if x≥2f(x) = \begin{cases} 2x - 3 & \text{if } x < -1 \\ x^2 + 2 & \text{if } -1 \leq x < 2 \\ 5x - 1 & \text{if } x \geq 2 \end{cases}f(x)=⎩⎨⎧​2x−3x2+25x−1​if x<−1if −1≤x<2if x≥2​, what is f(2)?

  1. 10
  2. 4
  3. 9 (correct answer)
  4. 3

Explanation: For x = 2, check intervals: 2<−12 < -12<−1? No. −1≤2<2-1 \leq 2 < 2−1≤2<2? No. 2≥22 \geq 22≥2? Yes. So use the third piece f(x)=5x−1f(x) = 5x - 1f(x)=5x−1. Substitute x = 2: f(2)=5(2)−1=10−1=9f(2) = 5(2) - 1 = 10 - 1 = 9f(2)=5(2)−1=10−1=9. Note that x = 2 falls in the third piece due to the ≥\geq≥ condition.

Question 8

For the piecewise function f(x)={3x2if x<−1x−5if −1≤x<22x+3if x≥2f(x) = \begin{cases} 3x^2 & \text{if } x < -1 \\ x - 5 & \text{if } -1 \leq x < 2 \\ 2x + 3 & \text{if } x \geq 2 \end{cases}f(x)=⎩⎨⎧​3x2x−52x+3​if x<−1if −1≤x<2if x≥2​, what is f(2)?

  1. 9
  2. 8
  3. 6
  4. 7 (correct answer)

Explanation: For x = 2, check intervals: 2 < -1? No. -1 ≤ 2 < 2? No. 2 ≥ 2? Yes. So use the third piece f(x)=2x+3f(x) = 2x + 3f(x)=2x+3. Substitute x = 2: f(2)=2(2)+3=4+3=7f(2) = 2(2) + 3 = 4 + 3 = 7f(2)=2(2)+3=4+3=7. Note that x = 2 falls in the third piece due to the ≥ condition.

Question 9

What is f(2) for the piecewise function: f(x)={−x+3if x<14xif 1≤x<3x2−1if x≥3f(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}f(x)=⎩⎨⎧​−x+34xx2−1​if x<1if 1≤x<3if x≥3​?

  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) = 4xf(x)=4x. Substitute x = 2: f(2)=4(2)=8f(2) = 4(2) = 8f(2)=4(2)=8. The value x = 2 falls clearly within the middle interval.

Question 10

A company assigns a performance rating f(x)f(x)f(x) based on an employee’s score xxx. 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 $x=0$?
  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.