For $f(x) = 2x + 1$ and $g(x) = x^2$, find $f(g(3))$ and $g(f(3))$.

$f(g(3)) = 19$ and $g(f(3)) = 49$

For $f(x) = x - 4$ and $g(x) = 3x^2$, find $f(g(x))$ and $g(f(x))$.

$f(g(x)) = 3x^2 - 4$; $g(f(x)) = 3x^2 - 24x + 48$

For $f(x) = \sqrt{x - 2}$ and $g(x) = x^2 + 1$, find $f(g(x))$ and $g(f(x))$ with their domains.

$f(g(x)) = \sqrt{x^2 - 1}$ for $x \le -1$ or $x \ge 1$; $g(f(x)) = x - 1$ for $x \ge 2$

For $f(x) = \frac{1}{x - 3}$ and $g(x) = 2x + 1$, find $f(g(x))$ and $g(f(x))$ and state each domain.

$f(g(x)) = \frac{1}{2x - 2}$ for $x \ne 1$; $g(f(x)) = \frac{x - 1}{x - 3}$ for $x \ne 3$

$f(x) = 2x - 3$ and $f(g(x)) = 6x^2 + 4x - 1$. Find $g(x)$, then evaluate $g(f(2))$.

$g(x) = 3x^2 + 2x + 1$; $g(f(2)) = 6$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Composition of Functions

A composite function feeds the output of one function into another. $f(g(x))$ , also written $(f \circ g)(x)$ , means: apply $g$ to $x$ first, then apply $f$ to the result. Read it from the inside out.

How It Works

Evaluating at a number. Work from the inside: to find $f(g(3))$ , find $g(3)$ , then put that number into $f$ . You do not need the composite equation for this.

Finding the composite equation. Replace every $x$ in the outer function with the whole inner function, in brackets. For $f(x) = x - 4$ and $g(x) = 3x^2$ , $f(g(x)) = 3x^2 - 4$ , but $g(f(x)) = 3(x - 4)^2$ . These are different functions: in general $f(g(x)) \ne g(f(x))$ , so the order matters.

The domain of a composite. $x$ has to be in the domain of the inner function, and the inner function's output has to be in the domain of the outer one. Simplifying can hide a restriction. With $f(x) = \sqrt{x - 2}$ and $g(x) = x^2 + 1$ , $g(f(x))$ simplifies to $x - 1$ , which looks defined everywhere, but $f$ only accepts $x \ge 2$ , so the composite does too.

Decomposing. Going the other way, a complicated function can often be written as a composite of two simple ones. $h(x) = (3x - 5)^4$ is $f(g(x))$ with $g(x) = 3x - 5$ inside and $f(x) = x^4$ outside. When one of $f$ and $f(g(x))$ is given and the other unknown, solve for the unknown inside function algebraically.

Inverses. A function and its inverse undo each other: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ . This is a quick check of an inverse you have found.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
For $f(x) = 2x + 1$ and $g(x) = x^2$ , find $f(g(3))$ and $g(f(3))$ .
Show solution
1. $g(3) = 9$ , so $f(g(3)) = f(9) = 19$ .
2. $f(3) = 7$ , so $g(f(3)) = g(7) = 49$ .
Answer: $f(g(3)) = 19$ and $g(f(3)) = 49$
Easy
For $f(x) = x - 4$ and $g(x) = 3x^2$ , find $f(g(x))$ and $g(f(x))$ .
Show solution
1. $f(g(x)) = f(3x^2) = 3x^2 - 4$ .
2. $g(f(x)) = g(x - 4) = 3(x - 4)^2 = 3x^2 - 24x + 48$ .
Answer: $f(g(x)) = 3x^2 - 4$ ; $g(f(x)) = 3x^2 - 24x + 48$
Medium
For $f(x) = \sqrt{x - 2}$ and $g(x) = x^2 + 1$ , find $f(g(x))$ and $g(f(x))$ with their domains.
Show solution
1. $f(g(x)) = \sqrt{x^2 + 1 - 2} = \sqrt{x^2 - 1}$ . It needs $x^2 \ge 1$ , so $x \le -1$ or $x \ge 1$ .
2. $g(f(x)) = (\sqrt{x - 2})^2 + 1 = x - 1$ .
3. But $f$ is applied first, so $x \ge 2$ . The domain is $x \ge 2$ , not all real numbers.
Answer: $f(g(x)) = \sqrt{x^2 - 1}$ for $x \le -1$ or $x \ge 1$ ; $g(f(x)) = x - 1$ for $x \ge 2$
Medium
For $f(x) = \frac{1}{x - 3}$ and $g(x) = 2x + 1$ , find $f(g(x))$ and $g(f(x))$ and state each domain.
Show solution
1. $f(g(x)) = \frac{1}{2x + 1 - 3} = \frac{1}{2x - 2}$ . The denominator is $0$ at $x = 1$ .
2. $g(f(x)) = \frac{2}{x - 3} + 1 = \frac{2 + x - 3}{x - 3} = \frac{x - 1}{x - 3}$ .
3. $f$ is applied first and excludes $x = 3$ .
Answer: $f(g(x)) = \frac{1}{2x - 2}$ for $x \ne 1$ ; $g(f(x)) = \frac{x - 1}{x - 3}$ for $x \ne 3$
Test level
$f(x) = 2x - 3$ and $f(g(x)) = 6x^2 + 4x - 1$ . Find $g(x)$ , then evaluate $g(f(2))$ .
Show solution
1. $f(g(x)) = 2g(x) - 3$ , so $2g(x) - 3 = 6x^2 + 4x - 1$ .
2. $2g(x) = 6x^2 + 4x + 2$ , so $g(x) = 3x^2 + 2x + 1$ .
3. Check: $2(3x^2 + 2x + 1) - 3 = 6x^2 + 4x - 1$ .
4. $f(2) = 1$ , so $g(f(2)) = g(1) = 3 + 2 + 1 = 6$ .
Answer: $g(x) = 3x^2 + 2x + 1$ ; $g(f(2)) = 6$

Learn It on Martian Lab

Martian Lab teaches this in the Composition of Functions lesson of its MHF4U Combining Functions unit. Watch it, then take the Combining Functions practice tests to see if it stuck.

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More Combining Functions topics: Sums, Products and Quotients of Functions · Average vs Instantaneous Rate of Change

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems

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