Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Piecewise Functions - ACT Math

  1. Based on the piecewise function f(x)={4x−1,x<5(x−5)2,5≤x<820−x,x≥8f(x)=\begin{cases} 4x-1, & x<5\\ (x-5)^2, & 5\le x<8\\ 20-x, & x\ge 8 \end{cases} what is the value when x=6x=6?
    Answer: 1 For x = 6, check which interval contains this value: 6<56 < 5? No. 5≤6<85 \le 6 < 8? Yes. Since x = 6 satisfies the condition 5≤x<85 \le x < 8, we use the second piece f(x)=(x−5)2f(x) = (x - 5)^2. Substituting x = 6: f(6)=(6−5)2=12=1f(6) = (6 - 5)^2 = 1^2 = 1. Choice A might result from incorrectly using the first piece where 4x−14x - 1 would give 4(6)−1=234(6) - 1 = 23.
  2. Based on the piecewise function f(x)={2−x,x<−4(x+4)(x−1),−4≤x<13,x≥1f(x)=\begin{cases} 2-x, & x<-4\\ (x+4)(x-1), & -4\le x<1\\ 3, & x\ge 1 \end{cases} what is the value when x=−4x=-4?
    Answer: 0 For x=−4x = -4, check which interval contains this value: −4<−4-4 < -4? No. −4≤−4<1-4 \leq -4 < 1? Yes. Since x=−4x = -4 satisfies the condition −4≤x<1-4 \leq x < 1, we use the second piece f(x)=(x+4)(x−1)f(x) = (x + 4)(x - 1). Substituting x=−4x = -4: f(−4)=(−4+4)(−4−1)=(0)(−5)=0f(-4) = (-4 + 4)(-4 - 1) = (0)(-5) = 0. The boundary point x=−4x = -4 is included in the second piece due to the ≤\leq symbol.
  3. For the function f(x)={x2+2xif x<12x−5if x≥1f(x) = \begin{cases} x^2 + 2x & \text{if } x < 1 \\ 2x - 5 & \text{if } x \geq 1 \end{cases}, what is f(1)f(1)?
    Answer: -3 For x = 1, we check the intervals: Is 1 < 1? No. Is 1 ≥ 1? Yes. Since x = 1 satisfies the second condition, we use f(x) = 2x - 5. Substituting x = 1: f(1) = 2(1) - 5 = 2 - 5 = -3. The boundary point x = 1 belongs to the second piece due to the ≥ symbol.
  4. For the function f(x)={2x+5if x<1x2if x≥1f(x) = \begin{cases} 2x + 5 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases}, what is f(0)f(0)?
    Answer: 5 For x = 0, we check which interval contains it: Is 0 < 1? Yes. So we use the first piece: f(x)=2x+5f(x) = 2x + 5. Substituting x = 0: f(0)=2(0)+5=0+5=5f(0) = 2(0) + 5 = 0 + 5 = 5. Choice B would result from using the second piece incorrectly.
  5. What is f(0)f(0) for the piecewise function defined below? f(x)={4x,x<0(x−2)2,0≤x<52x−3,x≥5f(x)=\begin{cases} 4x, & x<0\\ (x-2)^2, & 0\le x<5\\ 2x-3, & x\ge 5 \end{cases}
    Answer: 4 For x = 0, check which interval contains this value: 0 < 0? No. 0 ≤ 0 < 5? Yes. Since x = 0 satisfies the condition 0 ≤ x < 5, we use the second piece f(x) = (x - 2)². Substituting x = 0: f(0) = (0 - 2)² = (-2)² = 4. The boundary point x = 0 is included in the second piece due to the ≤ symbol.
  6. 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}?
    Answer: -14 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.
  7. 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)?
    Answer: 71 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.
  8. For the function f(x)={x2+1if x<03x−4if 0≤x<32x+5if x≥3f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x - 4 & \text{if } 0 \leq x < 3 \\ 2x + 5 & \text{if } x \geq 3 \end{cases}, what is f(-1)?
    Answer: 2 For x=−1x = -1, check intervals: −1<0-1 < 0? Yes. So use the first piece f(x)=x2+1f(x) = x^2 + 1. Substitute x=−1x = -1: f(−1)=(−1)2+1=1+1=2f(-1) = (-1)^2 + 1 = 1 + 1 = 2. Since -1 is negative, it falls in the first interval.
  9. For the piecewise function f(x)={(x+2)2,x<−25,−2≤x<12x−1,x≥1f(x)=\begin{cases} (x+2)^2, & x<-2\\ 5, & -2\le x<1\\ 2x-1, & x\ge 1 \end{cases} what is f(−2)f(-2)?
    Answer: 5 For x = -2, check which interval contains this value: −2<−2-2 < -2? No. −2≤−2<1-2 \le -2 < 1? Yes. Since x = -2 satisfies the condition −2≤x<1-2 \le x < 1, we use the second piece f(x) = 5. Substituting x = -2: f(−2)=5f(-2) = 5. The boundary point x = -2 is included in the second piece due to the ≤\le symbol, giving a constant value of 5.
  10. A sensor converts a reading xx into an alert level f(x)f(x) using the piecewise function below. For the function f(x)={3xif x<1x2+2if 1≤x<414−xif x≥4f(x)=\begin{cases}3x & \text{if } x<1\\ x^2+2 & \text{if } 1\le x<4\\ 14-x & \text{if } x\ge 4\end{cases} what is f(0)f(0)?
    Answer: 0 For x = 0, check intervals: Is 0 < 1? Yes. So we use the first piece: f(x) = 3x. Substituting x = 0: f(0) = 3(0) = 0. Since 0 is less than 1, we don't need to check other intervals.
  11. For the function f(x)={2x2,x<16−x,1≤x<4x+22,x≥4f(x)=\begin{cases} 2x^2, & x<1\\ 6-x, & 1\le x<4\\ \dfrac{x+2}{2}, & x\ge 4 \end{cases} what is f(4)f(4)?
    Answer: 3 For x=4x = 4, check which interval contains this value: 4<14 < 1? No. 1≤4<41 \le 4 < 4? No. 4≥44 \ge 4? Yes. Since x=4x = 4 satisfies the condition x≥4x \ge 4, we use the third piece f(x)=x+22f(x) = \dfrac{x+2}{2}. Substituting x=4x = 4: f(4)=4+22=62=3f(4) = \dfrac{4 + 2}{2} = \dfrac{6}{2} = 3. The boundary point x=4x = 4 falls into the third piece due to the ≥\ge symbol.
  12. 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}?
    Answer: 3 For x = 1, check intervals: 1 < 1? No. 1 ≤ 1 < 4? Yes. So use the second piece f(x)=3f(x) = 3. At the boundary x = 1, we use the second piece due to the ≤ condition. Therefore f(1)=3f(1) = 3.
  13. For the function f(x)={x+2if x≤−33x−7if x>−3f(x) = \begin{cases} x + 2 & \text{if } x \leq -3 \\ 3x - 7 & \text{if } x > -3 \end{cases}, what is f(−3)f(-3)?
    Answer: -1 For x=−3x = -3, we check which interval contains it: Is −3≤−3-3 \leq -3? Yes. So we use the first piece: f(x)=x+2f(x) = x + 2. Substituting x=−3x = -3: f(−3)=(−3)+2=−1f(-3) = (-3) + 2 = -1. The boundary point x=−3x = -3 is included in the first piece due to the ≤\leq symbol.
  14. For the function f(x)={4x−3if x<0x2+2if x≥0f(x) = \begin{cases} 4x - 3 & \text{if } x < 0 \\ x^2 + 2 & \text{if } x \geq 0 \end{cases}, what is f(−1)f(-1)?
    Answer: -7 For x = -1, we check which interval applies: Is -1 < 0? Yes. So we use the first piece: f(x)=4x−3f(x) = 4x - 3. Substituting x = -1: f(−1)=4(−1)−3=−4−3=−7f(-1) = 4(-1) - 3 = -4 - 3 = -7. Choice A would result from using the second piece incorrectly.
  15. Based on the piecewise function f(x)={(x+1)2−1,x<−12x+5,−1≤x<29−x,x≥2f(x)=\begin{cases} (x+1)^2-1, & x<-1\\ 2x+5, & -1\le x<2\\ 9-x, & x\ge 2 \end{cases} what is the value when x=−1x=-1?
    Answer: 3 For x=−1x = -1, check which interval contains this value: −1<−1-1 < -1? No. −1≤−1<2-1 \leq -1 < 2? Yes. Since x=−1x = -1 satisfies the condition −1≤x<2-1 \leq x < 2, we use the second piece f(x)=2x+5f(x) = 2x + 5. Substituting x=−1x = -1: f(−1)=2(−1)+5=−2+5=3f(-1) = 2(-1) + 5 = -2 + 5 = 3. The boundary point x=−1x = -1 is included in the second piece due to the ≤\leq symbol.
  16. For the function f(x)={4x+1if x<0x−3if 0≤x<52x2if x≥5f(x) = \begin{cases} 4x + 1 & \text{if } x < 0 \\ x - 3 & \text{if } 0 \leq x < 5 \\ 2x^2 & \text{if } x \geq 5 \end{cases}, what is f(7)?
    Answer: 98 For x = 7, check intervals: 7 < 0? No. 0 ≤ 7 < 5? No. 7 ≥ 5? Yes. So use the third piece f(x) = 2x22x^2. Substitute x = 7: f(7) = 2(7)2=2(49)=982(7)^2 = 2(49) = 98. The value falls clearly in the rightmost interval.
  17. A taxi company models a fare adjustment f(x)f(x) based on the distance xx (in miles). The adjustment is f(x)={4if x<22x+1if 2≤x<6x2−10if x≥6f(x)=\begin{cases} 4 & \text{if } x<2 \\ 2x+1 & \text{if } 2\le x<6 \\ x^2-10 & \text{if } x\ge 6 \end{cases} What is f(6)f(6) for the piecewise function defined below?
    Answer: 26 To find f(6), check intervals: Is 6 < 2? No. Is 2 ≤ 6 < 6? No, since 6 is not less than 6. Is 6 ≥ 6? Yes. Use the third piece: f(x) = x² - 10. Thus f(6) = 6² - 10 = 36 - 10 = 26. The boundary x = 6 belongs to the third piece.
  18. What is f(0)f(0) for the piecewise function f(x)={−3x+4if x<0x2if x≥0f(x) = \begin{cases} -3x + 4 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}?
    Answer: 0 For x = 0, we check the intervals: Is 0 < 0? No. Is 0 ≥ 0? Yes. Since x = 0 satisfies the second condition, we use f(x)=x2f(x) = x^2. Substituting x = 0: f(0)=(0)2=0f(0) = (0)^2 = 0. The boundary point x = 0 belongs to the second piece due to the ≥ symbol.
  19. A gym records a membership score using a piecewise function based on the number of visits xx in a month. The function is f(x)={−x+12if x<42x−1if 4≤x<920if x≥9f(x)=\begin{cases} -x+12 & \text{if } x<4 \\ 2x-1 & \text{if } 4\le x<9 \\ 20 & \text{if } x\ge 9 \end{cases} Based on the piecewise function, what is the value when x=8x=8?
    Answer: 15 For x = 8, check intervals: Is 8 < 4? No. Is 4 ≤ 8 < 9? Yes, since 8 is between 4 and 9. Use the second piece: f(x) = 2x - 1. Substituting: f(8) = 2(8) - 1 = 16 - 1 = 15. Choice C (20) incorrectly used the third piece for x ≥ 9.
  20. 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}, what is f(3)?
    Answer: 4 For x=3x = 3, check intervals: 3<−23 < -2? No. −2≤3<3-2 \leq 3 < 3? No. 3≥33 \geq 3? Yes. So use the third piece f(x)=x2−5f(x) = x^2 - 5. Substitute x=3x = 3: f(3)=32−5=9−5=4f(3) = 3^2 - 5 = 9 - 5 = 4. Note that x=3x = 3 falls in the third piece due to the ≥\geq condition.

Start this quiz