Piecewise function composition (ACT). Game on.

Piecewise function composition (ACT) is a grade 9 math skill aligned to Common Core standard HSF.BF.A.1.c. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 piecewise function composition (act) problems our math games drill.

CCSS HSF.BF.A.1.c10 questions in the bank
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Warm-upmedium

Use the piecewise definitions of ff and gg.
f(x)={x+3for x≤32x−1for x>3f(x)=\begin{cases} x+3 & \text{for } x \leq 3 \\ 2x-1 & \text{for } x > 3 \end{cases}
g(x)={4−xfor x≤1x+2for x>1g(x)=\begin{cases} 4-x & \text{for } x \leq 1 \\ x+2 & \text{for } x > 1 \end{cases}
What is
f(g(−1))?f(g(-1))\textsf{?}

Start with the inner function: −1≤1-1 \leq 1, so g(−1)=4−(−1)=5g(-1)=4-(-1)=5. Then 5>35 > 3, so f(5)=2(5)−1=9.f(5)=2(5)-1=9\textsf{.}

Mid-gamemedium

ff and gg are piecewise functions.
f(x)={5−xfor x≤42xfor x>4f(x)=\begin{cases} 5-x & \text{for } x \leq 4 \\ 2x & \text{for } x > 4 \end{cases}
g(x)={x+6for x≤03xfor x>0g(x)=\begin{cases} x+6 & \text{for } x \leq 0 \\ 3x & \text{for } x > 0 \end{cases}
Evaluate
f(g(0)).f(g(0))\textsf{.}

Start with the inner function: 0≤00 \leq 0, so g(0)=0+6=6g(0)=0+6=6. Then 6>46 > 4, so f(6)=2(6)=12.f(6)=2(6)=12\textsf{.}

Mid-gamemedium

The functions ff and gg are defined piecewise as follows.
f(x)={∣x∣−1for x≤23xfor x>2f(x)=\begin{cases} |x|-1 & \text{for } x \leq 2 \\ 3x & \text{for } x > 2 \end{cases}
g(x)={x2for x<14−xfor x≥1g(x)=\begin{cases} x^{2} & \text{for } x < 1 \\ 4-x & \text{for } x \geq 1 \end{cases}
What is the value of
g(f(−4))?g(f(-4))\textsf{?}

Start with the inner function: −4≤2-4 \leq 2, so f(−4)=∣−4∣−1=3f(-4)=|-4|-1=3. Then 3≥13 \geq 1, so g(3)=4−3=1.g(3)=4-3=1\textsf{.}

Mid-gamemedium

ff and gg are defined by
f(x)={3xfor x<2x+1for x≥2f(x)=\begin{cases} 3x & \text{for } x < 2 \\ x+1 & \text{for } x \geq 2 \end{cases}
g(x)={x−4for x≤52xfor x>5g(x)=\begin{cases} x-4 & \text{for } x \leq 5 \\ 2x & \text{for } x > 5 \end{cases}
What is
f(g(5))?f(g(5))\textsf{?}

Start with the inner function: 5≤55 \leq 5, so g(5)=5−4=1g(5)=5-4=1. Then 1<21 < 2, so f(1)=3(1)=3.f(1)=3(1)=3\textsf{.}

Mid-gamemedium

Consider the piecewise functions
f(x)={2−xfor x≤0x+5for x>0f(x)=\begin{cases} 2-x & \text{for } x \leq 0 \\ x+5 & \text{for } x > 0 \end{cases}
g(x)={−2xfor x<3x−1for x≥3g(x)=\begin{cases} -2x & \text{for } x < 3 \\ x-1 & \text{for } x \geq 3 \end{cases}
The value of
f(g(3))f(g(3)) is

Start with the inner function: 3≥33 \geq 3, so g(3)=3−1=2g(3)=3-1=2. Then 2>02 > 0, so f(2)=2+5=7.f(2)=2+5=7\textsf{.}

Mid-gamemedium

Piecewise functions ff and gg are shown.
f(x)={∣x∣+3for x≤42x−5for x>4f(x)=\begin{cases} |x|+3 & \text{for } x \leq 4 \\ 2x-5 & \text{for } x > 4 \end{cases}
g(x)={6−xfor x≤0x+1for x>0g(x)=\begin{cases} 6-x & \text{for } x \leq 0 \\ x+1 & \text{for } x > 0 \end{cases}
Find the value of
f(g(0)).f(g(0))\textsf{.}

Start with the inner function: 0≤00 \leq 0, so g(0)=6−0=6g(0)=6-0=6. Then 6>46 > 4, so f(6)=2(6)−5=7.f(6)=2(6)-5=7\textsf{.}

Mid-gamehard

The piecewise functions ff and gg are given.
f(x)={4x−1for x≤−13−xfor x>−1f(x)=\begin{cases} 4x-1 & \text{for } x \leq -1 \\ 3-x & \text{for } x > -1 \end{cases}
g(x)={x+8for x≤25xfor x>2g(x)=\begin{cases} x+8 & \text{for } x \leq 2 \\ 5x & \text{for } x > 2 \end{cases}
f(g(1))=f(g(1))=

Start with the inner function: 1≤21 \leq 2, so g(1)=1+8=9g(1)=1+8=9. Then 9>−19 > -1, so f(9)=3−9=−6.f(9)=3-9=-6\textsf{.}

Buzzer beaterhard

Using these definitions of ff and gg,
f(x)={x+2for x<01−2xfor x≥0f(x)=\begin{cases} x+2 & \text{for } x < 0 \\ 1-2x & \text{for } x \geq 0 \end{cases}
g(x)={3xfor x≤1x2for x>1g(x)=\begin{cases} 3x & \text{for } x \leq 1 \\ x^{2} & \text{for } x > 1 \end{cases}
what is
g(f(4))?g(f(4))\textsf{?}

Start with the inner function: 4≥04 \geq 0, so f(4)=1−2(4)=−7f(4)=1-2(4)=-7. Then −7≤1-7 \leq 1, so g(−7)=3(−7)=−21.g(-7)=3(-7)=-21\textsf{.}

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