Function table composition (ACT). Game on.

Function table 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 function table composition (act) problems our math games drill.

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

The table defines functions for positive integers x5x \leq 5.
xm(x)n(x)k(x)14312254351241455323\begin{array}{c|c|c|c} x & m(x) & n(x) & k(x) \\ \hline 1 & 4 & 3 & 1 \\ 2 & 2 & 5 & 4 \\ 3 & 5 & 1 & 2 \\ 4 & 1 & 4 & 5 \\ 5 & 3 & 2 & 3 \end{array}
If
m(n(k(x)))=1m(n(k(x))) = 1, what is the value of xx?

Work from the outside in. Since m(4)=1m(4) = 1, we need n(k(x))=4n(k(x)) = 4. Since n(4)=4n(4) = 4, we need k(x)=4k(x) = 4. From the table, k(2)=4k(2) = 4, so x=2x = 2.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xa(x)b(x)c(x)15242143335142155432\begin{array}{c|c|c|c} x & a(x) & b(x) & c(x) \\ \hline 1 & 5 & 2 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 3 & 5 & 1 \\ 4 & 2 & 1 & 5 \\ 5 & 4 & 3 & 2 \end{array}
If
a(b(c(x)))=4a(b(c(x))) = 4, what is the value of xx?

Work from the outside in. Since a(5)=4a(5) = 4, we need b(c(x))=5b(c(x)) = 5. Since b(3)=5b(3) = 5, we need c(x)=3c(x) = 3. From the table, c(2)=3c(2) = 3, so x=2x = 2.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xp(x)q(x)r(x)13152542323444515123\begin{array}{c|c|c|c} x & p(x) & q(x) & r(x) \\ \hline 1 & 3 & 1 & 5 \\ 2 & 5 & 4 & 2 \\ 3 & 2 & 3 & 4 \\ 4 & 4 & 5 & 1 \\ 5 & 1 & 2 & 3 \end{array}
If
p(q(r(x)))=1p(q(r(x))) = 1, what is the value of xx?

Work from the outside in. Since p(5)=1p(5) = 1, we need q(r(x))=5q(r(x)) = 5. Since q(4)=5q(4) = 5, we need r(x)=4r(x) = 4. From the table, r(3)=4r(3) = 4, so x=3x = 3.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xm(x)n(x)k(x)12432512345143255134\begin{array}{c|c|c|c} x & m(x) & n(x) & k(x) \\ \hline 1 & 2 & 4 & 3 \\ 2 & 5 & 1 & 2 \\ 3 & 4 & 5 & 1 \\ 4 & 3 & 2 & 5 \\ 5 & 1 & 3 & 4 \end{array}
If
m(n(k(x)))=3m(n(k(x))) = 3, what is the value of xx?

Work from the outside in. Since m(4)=3m(4) = 3, we need n(k(x))=4n(k(x)) = 4. Since n(1)=4n(1) = 4, we need k(x)=1k(x) = 1. From the table, k(3)=1k(3) = 1, so x=3x = 3.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xu(x)v(x)w(x)11352524324144535312\begin{array}{c|c|c|c} x & u(x) & v(x) & w(x) \\ \hline 1 & 1 & 3 & 5 \\ 2 & 5 & 2 & 4 \\ 3 & 2 & 4 & 1 \\ 4 & 4 & 5 & 3 \\ 5 & 3 & 1 & 2 \end{array}
If
u(v(w(x)))=4u(v(w(x))) = 4, what is the value of xx?

Work from the outside in. Since u(4)=4u(4) = 4, we need v(w(x))=4v(w(x)) = 4. Since v(3)=4v(3) = 4, we need w(x)=3w(x) = 3. From the table, w(4)=3w(4) = 3, so x=4x = 4.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xp(x)q(x)r(x)14512235314245235314\begin{array}{c|c|c|c} x & p(x) & q(x) & r(x) \\ \hline 1 & 4 & 5 & 1 \\ 2 & 2 & 3 & 5 \\ 3 & 1 & 4 & 2 \\ 4 & 5 & 2 & 3 \\ 5 & 3 & 1 & 4 \end{array}
If
p(q(r(x)))=5p(q(r(x))) = 5, what is the value of xx?

Work from the outside in. Since p(4)=5p(4) = 5, we need q(r(x))=4q(r(x)) = 4. Since q(3)=4q(3) = 4, we need r(x)=3r(x) = 3. From the table, r(4)=3r(4) = 3, so x=4x = 4.

Mid-gamemedium

The table defines functions for positive integers x5x \leq 5.
xm(x)n(x)k(x)15242341313544525213\begin{array}{c|c|c|c} x & m(x) & n(x) & k(x) \\ \hline 1 & 5 & 2 & 4 \\ 2 & 3 & 4 & 1 \\ 3 & 1 & 3 & 5 \\ 4 & 4 & 5 & 2 \\ 5 & 2 & 1 & 3 \end{array}
If
m(n(k(x)))=1m(n(k(x))) = 1, what is the value of xx?

Work from the outside in. Since m(3)=1m(3) = 1, we need n(k(x))=3n(k(x)) = 3. Since n(3)=3n(3) = 3, we need k(x)=3k(x) = 3. From the table, k(5)=3k(5) = 3, so x=5x = 5.

Buzzer beatermedium

The table defines functions for positive integers x5x \leq 5.
xj(x)k(x)l(x)13422153342545145231\begin{array}{c|c|c|c} x & j(x) & k(x) & l(x) \\ \hline 1 & 3 & 4 & 2 \\ 2 & 1 & 5 & 3 \\ 3 & 4 & 2 & 5 \\ 4 & 5 & 1 & 4 \\ 5 & 2 & 3 & 1 \end{array}
If
j(k(l(x)))=5j(k(l(x))) = 5, what is the value of xx?

Work from the outside in. Since j(4)=5j(4) = 5, we need k(l(x))=4k(l(x)) = 4. Since k(1)=4k(1) = 4, we need l(x)=1l(x) = 1. From the table, l(5)=1l(5) = 1, so x=5x = 5.

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