Math Core

A functional equation describes a function by a rule it must obey, like $f(x + y) = f(x) + f(y)$ , instead of by a formula. AMC problems of this kind almost never need heavy theory. The winning moves are to plug in clever values, to create a second equation and solve a system, and to spot patterns when a function is applied over and over.

Plug in special values

Try inputs that make the equation collapse: $0$ , $1$ , $-1$ , or setting two variables equal. Each substitution is a new fact about $f$ .

Worked example: Building up values

A function satisfies $f(x + y) = f(x) + f(y) + 2xy$ for all real $x, y$ , and $f(1) = 3$ . Find $f(5)$ .

Set $x = y = 0$ : $f(0) = 2f(0)$ , so $f(0) = 0$ . Set $y = 1$ : $f(x + 1) = f(x) + 3 + 2x$ . Now step up:

f(2) = 3 + 3 + 2 = 8,\quad f(3) = 8 + 3 + 4 = 15,\quad f(4) = 15 + 3 + 6 = 24,\quad f(5) = 24 + 3 + 8 = 35.

You can also guess the form. The $2xy$ term suggests $x^2$ , and $f(x) = x^2 + cx$ satisfies the equation for any $c$ . From $f(1) = 1 + c = 3$ , $c = 2$ , and $f(5) = 25 + 10 = 35$ .

Make a second equation

If the equation mixes $f(x)$ with $f(\text{something else})$ , such as $f\!\left(\frac{1}{x}\right)$ , $f(-x)$ or $f(1 - x)$ , substitute that "something else" for $x$ . You get a second equation with the same two unknowns, and you solve the $2 \times 2$ linear system.

The swap substitution

If an equation relates $f(x)$ and $f(g(x))$ , where applying $g$ twice gets you back to $x$ , then replacing $x$ with $g(x)$ gives a second equation in the same two unknowns. Common swaps:

x \to \frac{1}{x}, \qquad x \to -x, \qquad x \to 1 - x.

Treat $f(x)$ and $f(g(x))$ like the variables in a linear system and eliminate one.

Worked example: A reciprocal swap

A function satisfies $f(x) + 2f\!\left(\dfrac{1}{x}\right) = 3x$ for all $x \ne 0$ . Find $f(2)$ .

Plug in $x = 2$ and $x = \frac{1}{2}$ :

\begin{aligned} f(2) + 2f\!\left(\tfrac{1}{2}\right) &= 6 \\ f\!\left(\tfrac{1}{2}\right) + 2f(2) &= \tfrac{3}{2} \end{aligned}

From the second, $f\!\left(\frac{1}{2}\right) = \frac{3}{2} - 2f(2)$ . Substitute into the first: $f(2) + 3 - 4f(2) = 6$ , so $f(2) = -1$ . Check: $f\!\left(\frac{1}{2}\right) = \frac{7}{2}$ , and $-1 + 7 = 6$ . ✓

Iterate and look for a cycle

When a rule gives $f(x + 1)$ in terms of $f(x)$ , or a problem asks for $f(f(\cdots f(x)))$ many times, compute the first few values. They often repeat.

Worked example: A periodic function

A function satisfies $f(x + 1) = \dfrac{1 + f(x)}{1 - f(x)}$ and $f(0) = 2$ . Find $f(2026)$ .

Compute:

f(1) = \frac{3}{-1} = -3,\quad f(2) = \frac{-2}{4} = -\frac{1}{2},\quad f(3) = \frac{1/2}{3/2} = \frac{1}{3},\quad f(4) = \frac{4/3}{2/3} = 2.

The values repeat every $4$ steps. Since $2026 = 4 \cdot 506 + 2$ , $f(2026) = f(2) = -\dfrac{1}{2}$ .

Composition and inverses

To find $f^{-1}(a)$ , solve $f(x) = a$ ; you don't need the whole inverse formula. The graphs of $f$ and $f^{-1}$ are reflections of each other over the line $y = x$ .

y = 2x + 1 and its inverse y = (x − 1)/2 are mirror images over y = x.Open in grapher →

Worked example: Composition of linear functions

A linear function $f(x) = ax + b$ with $a > 0$ satisfies $f(f(x)) = 9x + 8$ . Find $f(1)$ .

$f(f(x)) = a(ax + b) + b = a^2 x + (ab + b)$ . Match coefficients: $a^2 = 9$ , so $a = 3$ , and $ab + b = 4b = 8$ , so $b = 2$ . Then $f(1) = 5$ .

Common mistake

Don't assume a function is linear (or any particular form) just because a guess fits one value. A guessed form is fine for finding a multiple-choice answer quickly, but check it satisfies the equation for all inputs, not just the ones you tried.

Tip

If a function is a sum of odd powers plus a constant, like $g(x) = ax^5 + bx^3 + cx + d$ , then $g(x) - d$ is an odd function, so $g(x) + g(-x) = 2d$ . That turns a value at $-x$ into a value at $x$ instantly.

Practice

Practice 1

A function satisfies $f(2x + 1) = 4x^2 + 2x - 3$ for all real $x$ . What is $f(5)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Let $f(x) = ax^5 + bx^3 + 2x + 4$ , where $a$ and $b$ are constants. If $f(-3) = 10$ , what is $f(3)$ ?

Practice 3

A function defined on the positive reals satisfies $f(xy) = f(x) + f(y)$ for all positive $x, y$ . If $f(2) = 3$ and $f(3) = 5$ , what is $f(72)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A function satisfies $f(x) + 3f(1 - x) = x^2$ for all real $x$ . What is $f(0)$ ?

Practice 5

Let $f(x) = \dfrac{1}{1 - x}$ . Define $f^{(1)}(x) = f(x)$ and $f^{(n+1)}(x) = f\!\left(f^{(n)}(x)\right)$ . What is $f^{(2026)}(3)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Let $f(x) = \dfrac{2x + 1}{x - 3}$ . What is $f^{-1}(5)$ ?

Practice 7

A sequence of values satisfies $f(1) = 3$ , $f(2) = 5$ , and $f(n + 2) = f(n + 1) - f(n)$ for every positive integer $n$ . What is $f(1) + f(2) + \dots + f(2026)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A function satisfies $2f(x) + f(1 - x) = x^2$ for all real $x$ . What is $f(4)$ ?

Real contest practice

2001 AMC 12, Problem 9: a multiplicative functional equation; plug in a special value.
2006 AMC 12A, Problem 18: a rule linking $f(x)$ and $f\!\left(\frac{1}{x}\right)$ ; which inputs are allowed?