Composition of Functions (gof and fog)

Reading the notation

$gof$ means "first apply $f$ , then apply $g$ to the result", read right to left: $(g \circ f)(x) = g(f(x))$ . Likewise $fog(x) = f(g(x))$ . There's no reason these have to match.

A quick numerical check: with $f(x) = x + 1$ and $g(x) = x^2$ , $gof(2) = g(3) = 9$ but $fog(2) = f(4) = 5$ .

When a composition is defined

For $f: A \to B$ and $g: B \to C$ , the composition $gof: A \to C$ is defined because every output of $f$ is a valid input for $g$ . In general, $gof$ exists only when the range of $f$ sits inside the domain of $g$ .

Example: take $f(x) = \sqrt{x}$ (defined for $x \geq 0$ ) and $g(x) = x - 4$ .

$gof(x) = \sqrt{x} - 4$ , defined for every $x \geq 0$ .
$fog(x) = \sqrt{x - 4}$ , defined only for $x \geq 4$ , because $f$ can't accept the negative outputs $g$ produces when $x < 4$ .

Same two functions, different order, different domain.

Worked example

$f,g: \mathbb{R} \to \mathbb{R}$ are given by $g(x) = x^{1/3}$ and $f(x) = 8x^3$ . Find $gof$ and $fog$ .

Solution, gof: substitute $f(x)$ into $g$ :

gof(x) = g(8x^3) = (8x^3)^{1/3} = 8^{1/3} \cdot x = 2x

fog: substitute $g(x)$ into $f$ :

fog(x) = f(x^{1/3}) = 8(x^{1/3})^3 = 8x

So $gof(x) = 2x$ but $fog(x) = 8x$ , a clean reminder that composition order matters.

Worked example: a function that undoes itself

If $f(x) = \dfrac{4x+3}{6x-4}$ for $x \neq \dfrac{2}{3}$ , show that $fof(x) = x$ .

Solution: Substitute $f(x)$ into itself, then multiply the top and bottom by $6x - 4$ :

fof(x) = \dfrac{4\left(\dfrac{4x+3}{6x-4}\right) + 3}{6\left(\dfrac{4x+3}{6x-4}\right) - 4} = \dfrac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} = \dfrac{34x}{34} = x

So $f$ is its own inverse: applying it twice gets you back where you started.

Checking an inverse with composition

A function $g$ is the inverse of $f$ exactly when both compositions give back the input: $gof(x) = x$ and $fog(x) = x$ .

Example: for $f(x) = 2x + 3$ , the candidate inverse is $g(x) = \dfrac{x - 3}{2}$ . Checking both orders:

gof(x) = \dfrac{(2x+3) - 3}{2} = x, \qquad fog(x) = 2\left(\dfrac{x-3}{2}\right) + 3 = x

Properties worth knowing

If	Then
$f$ and $g$ are both one-one	$gof$ is one-one
$f$ and $g$ are both onto	$gof$ is onto
$gof$ is one-one	$f$ is one-one (but $g$ need not be)
$gof$ is onto	$g$ is onto (but $f$ need not be)
Any three functions	$h \circ (g \circ f) = (h \circ g) \circ f$

Composition is associative, as the last row says, but not commutative: $gof$ and $fog$ usually differ, as the first two examples on this page show. They agree in special cases, such as a function and its own inverse.

Common mistakes

Reading $gof$ left to right. The function nearest $x$ acts first: in $g(f(x))$ that's $f$ .
Confusing composition with multiplication. $fog(x)$ is $f(g(x))$ , not $f(x) \cdot g(x)$ .
Ignoring the domain. Simplifying $fog(x) = \sqrt{x - 4}$ is only half the answer. State that it needs $x \geq 4$ .
Checking only one order for an inverse. For functions on restricted domains, $gof(x) = x$ can hold while $fog(x) = x$ fails. Check both.

Where this comes up in exams

The composition of functions is listed in the sets, relations and functions units of both the JEE Main 2026 and JEE Advanced 2026 syllabi.

Checked against JEE (Main) 2026 syllabus (NTA) and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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