Function Composition on the SAT
The one idea to hold onto
$f(g(x))$ means you do $g$ first and then put the result into $f$. You always work from the inside out. With a number, find $g(4)$, then find $f$ of that answer.
With an expression, replace every $x$ in $f$ with the whole of $g(x)$, in parentheses. The order matters: $f(g(x))$ and $g(f(x))$ are usually different, and the SAT likes to put both among the answer choices.
Composition questions are short. Almost every wrong answer I see comes from doing the outer function first, or from substituting into only part of the rule.
Examples
Both orders
$f(x) = 2x + 1$ and $g(x) = x^2$. Find $f(g(x))$ and $g(f(x))$.
- $f(g(x)) = 2(x^2) + 1 = 2x^2 + 1$.
- $g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1$.
- The two are different, which is why the SAT likes to offer both.
Finding the outer function
$g(x) = 2x + 1$ and $f(g(x)) = 6x + 5$, where $f$ is linear. Find $f(4)$.
- Write $6x + 5$ in terms of $2x + 1$: $6x + 5 = 3(2x + 1) + 2$.
- So $f(u) = 3u + 2$.
- $f(4) = 3(4) + 2 = 14$.
Solving an equation with a composition
$f(x) = x^2 - 4$ and $g(x) = x + 2$. For what values of $x$ is $f(g(x)) = 0$?
- $f(g(x)) = (x + 2)^2 - 4$.
- Set it to $0$: $(x + 2)^2 = 4$, so $x + 2 = 2$ or $x + 2 = -2$.
- $x = 0$ or $x = -4$.
What goes wrong
- Working outside in. In $f(g(4))$, $g$ goes first.
- Treating composition as multiplication. $f(g(x))$ is not $f(x) \cdot g(x)$.
- Leaving out parentheses. $g(f(x))$ with $g(x) = x^2$ is $(2x + 1)^2$, not $2x^2 + 1$.
- Picking the other order. The answer choices often include $g(f(x))$ when the question asks for $f(g(x))$.
- Squaring a binomial term by term. $(2x + 1)^2 = 4x^2 + 4x + 1$, not $4x^2 + 1$.
Quick reference
| You see | Do this |
|---|---|
| $f(g(4))$ | Find $g(4)$, then put that number into $f$. |
| $f(g(x))$ as an expression | Replace each $x$ in $f$ with $(g(x))$, then simplify. |
| $f(f(2))$ | Apply $f$ twice: find $f(2)$, then $f$ of that. |
| $f(g(x))$ given, $g$ given, find $f$ | Let $u = g(x)$ and rewrite the result in terms of $u$. |
| Domain of $f(g(x))$ | The inside output must be a value $f$ accepts, such as $\ge 0$ under a square root. |
| Tables or graphs of $f$ and $g$ | Read $g$ at the input first, then look up that output in $f$. |
Practice questions
Work from the inside out on every one. Then open the solution to compare.
Easy
If $f(x) = x + 3$ and $g(x) = 2x$, what is the value of $f(g(4))$?
- A$11$
- B$14$
- C$8$
- D$24$
Show solution
Answer: A. $g(4) = 8$, then $f(8) = 11$. Choice B is $g(f(4)) = g(7) = 14$, the other order.
EasyStudent-produced response
If $f(x) = x^2$ and $g(x) = x - 5$, what is the value of $g(f(3))$?
Enter your answer.
Show solution
Answer: $4$. $f(3) = 9$, then $g(9) = 9 - 5 = 4$.
Medium
If $f(x) = 2x + 1$ and $g(x) = x^2$, which expression is equal to $f(g(x))$?
- A$(2x + 1)^2$
- B$2x^2 + 1$
- C$4x^2 + 1$
- D$2x^3 + x^2$
Show solution
Answer: B. Replace $x$ in $f$ with $x^2$: $2(x^2) + 1 = 2x^2 + 1$. Choice A is $g(f(x))$, and choice D multiplies $f(x)$ by $g(x)$.
Medium
If $f(x) = 2x + 1$ and $g(x) = x^2$, which expression is equal to $g(f(x))$?
- A$2x^2 + 1$
- B$4x^2 + 1$
- C$4x^2 + 4x + 1$
- D$2x^2 + 4x + 1$
Show solution
Answer: C. $g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1$. Choice B squares each term separately and drops the middle term.
MediumStudent-produced response
If $f(x) = 3x - 2$, what is the value of $f(f(2))$?
Enter your answer.
Show solution
Answer: $10$. $f(2) = 4$, then $f(4) = 12 - 2 = 10$.
Medium
If $f(x) = 3x + 1$ and $g(x) = x - 4$, which expression is equal to $f(g(x))$?
- A$3x - 3$
- B$3x - 11$
- C$3x - 12$
- D$4x - 3$
Show solution
Answer: B. $f(g(x)) = 3(x - 4) + 1 = 3x - 12 + 1 = 3x - 11$. Choice A is $g(f(x))$, and choice C forgets the $+1$.
HardStudent-produced response
The function $f$ is linear, $g(x) = 2x + 1$, and $f(g(x)) = 6x + 5$ for all values of $x$. What is the value of $f(4)$?
Enter your answer.
Show solution
Answer: $14$. $6x + 5 = 3(2x + 1) + 2$, so $f(u) = 3u + 2$. Then $f(4) = 14$. Check: $f(g(x)) = 3(2x + 1) + 2 = 6x + 5$.
Hard
Let $f(x) = \sqrt{x}$ and $g(x) = x - 7$. For which values of $x$ is $f(g(x))$ defined as a real number?
- A$x \ge 0$
- B$x \ge -7$
- CAll real numbers
- D$x \ge 7$
Show solution
Answer: D. $f(g(x)) = \sqrt{x - 7}$, which needs $x - 7 \ge 0$, so $x \ge 7$. Choice A is the domain of $f$ alone.
Hard
Let $f(x) = x^2 - 4$ and $g(x) = x + 2$. What are all values of $x$ for which $f(g(x)) = 0$?
- A$0$ and $-4$
- B$2$ and $-2$
- C$-2$ only
- D$0$ and $4$
Show solution
Answer: A. $(x + 2)^2 - 4 = 0$ gives $x + 2 = \pm 2$, so $x = 0$ or $x = -4$. Choice B solves $f(x) = 0$ instead of $f(g(x)) = 0$.
HardStudent-produced response
If $f(x) = 2^x$ and $g(x) = x + 3$, what is the value of $g(f(2)) + f(g(-1))$?
Enter your answer.
Show solution
Answer: $11$. $g(f(2)) = g(4) = 7$. $f(g(-1)) = f(2) = 4$. The sum is $11$.
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