Composition of Functions

Master function composition, inverse functions, and their properties for JEE Mathematics.

5 min read

Introduction

When we apply one function after another, we create a composite function. Composition is a powerful way to build complex functions from simpler ones.

Composition of Functions

Definition

If $f: A \to B$ and $g: B \to C$, then the composition of $g$ with $f$, denoted $g \circ f$, is:

$$\boxed{(g \circ f)(x) = g(f(x))}$$

Read as: “$g$ of $f$ of $x$” or “$g$ composed with $f$”

Domain of Composition

$$\text{Dom}(g \circ f) = \{x \in A : f(x) \in \text{Dom}(g)\}$$

Visualization

    f         g
A ────► B ────► C

    g ∘ f
A ─────────────► C

Example 1

Let $f(x) = 2x + 1$ and $g(x) = x^2$.

$$(g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$$ $$(f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 1$$
Order Matters!

In general, $g \circ f \neq f \circ g$

Composition is NOT commutative!

Properties of Composition

1. Associativity

For $f: A \to B$, $g: B \to C$, $h: C \to D$:

$$\boxed{(h \circ g) \circ f = h \circ (g \circ f)}$$

Both equal $h(g(f(x)))$.

2. Identity Function

The identity function $I(x) = x$ acts as identity for composition:

$$f \circ I = I \circ f = f$$

3. Non-Commutativity

Generally: $f \circ g \neq g \circ f$

4. Injectivity and Surjectivity

IfThen
$f$ and $g$ are one-one$g \circ f$ is one-one
$f$ and $g$ are onto$g \circ f$ is onto
$f$ and $g$ are bijective$g \circ f$ is bijective

Converse results:

IfThen
$g \circ f$ is one-one$f$ is one-one
$g \circ f$ is onto$g$ is onto

Interactive Demo: Visualize Function Composition

Explore how composing functions affects their graphs.

Inverse Functions

Definition

For a bijective function $f: A \to B$, the inverse function $f^{-1}: B \to A$ satisfies:

$$\boxed{f^{-1}(y) = x \iff f(x) = y}$$

Key Properties

  1. $f^{-1} \circ f = I_A$ (identity on $A$)
  2. $f \circ f^{-1} = I_B$ (identity on $B$)
  3. $(f^{-1})^{-1} = f$
  4. $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ (reverses order!)

Finding the Inverse

Step 1: Write $y = f(x)$

Step 2: Solve for $x$ in terms of $y$

Step 3: Replace $y$ with $x$ to get $f^{-1}(x)$

Example 2

Find the inverse of $f(x) = 2x + 3$.

Solution:

  1. Let $y = 2x + 3$
  2. Solve for $x$: $x = \frac{y - 3}{2}$
  3. Replace $y$ with $x$: $f^{-1}(x) = \frac{x - 3}{2}$

Verification: $(f \circ f^{-1})(x) = f\left(\frac{x-3}{2}\right) = 2 \cdot \frac{x-3}{2} + 3 = x - 3 + 3 = x$ (verified)

Example 3

Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$, $x \neq 3$.

Solution:

  1. Let $y = \frac{2x + 1}{x - 3}$
  2. Cross multiply: $y(x - 3) = 2x + 1$
  3. Expand: $xy - 3y = 2x + 1$
  4. Collect $x$ terms: $xy - 2x = 3y + 1$
  5. Factor: $x(y - 2) = 3y + 1$
  6. Solve: $x = \frac{3y + 1}{y - 2}$
$$f^{-1}(x) = \frac{3x + 1}{x - 2}$$

Graph of Inverse Function

The graph of $f^{-1}$ is the reflection of the graph of $f$ about the line $y = x$.

If $(a, b)$ is on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$.

Self-Inverse Functions

A function $f$ is self-inverse (or an involution) if:

$$f^{-1} = f \quad \text{or} \quad f(f(x)) = x$$

Examples of self-inverse functions:

  • $f(x) = \frac{1}{x}$
  • $f(x) = -x$
  • $f(x) = \frac{a - x}{1 + ax}$ (for $ax \neq -1$)

Composition Examples

Example 4: Multiple Compositions

If $f(x) = x^2$ and $g(x) = x + 1$, find:

a) $(g \circ f)(3)$ b) $(f \circ g)(3)$ c) $(g \circ g)(3)$ d) $(f \circ f)(2)$

Solutions:

a) $(g \circ f)(3) = g(f(3)) = g(9) = 9 + 1 = 10$

b) $(f \circ g)(3) = f(g(3)) = f(4) = 16$

c) $(g \circ g)(3) = g(g(3)) = g(4) = 5$

d) $(f \circ f)(2) = f(f(2)) = f(4) = 16$

Example 5: Finding Component Functions

If $(g \circ f)(x) = \sin^2(x)$, find possible $f$ and $g$.

Solution:

One possibility:

  • $f(x) = \sin(x)$
  • $g(x) = x^2$

Then $(g \circ f)(x) = g(\sin(x)) = \sin^2(x)$ (correct)

Another possibility:

  • $f(x) = \sin^2(x)$
  • $g(x) = x$

There are infinitely many solutions!

JEE Important Results

If $f \circ g = I$ and $g \circ f = I$

Then $f$ and $g$ are inverses of each other.

If $f: \mathbb{R} \to \mathbb{R}$ is bijective

Then $f^{-1}$ exists and:

  • $(f^{-1})'(y) = \frac{1}{f'(x)}$ where $y = f(x)$

Periodic Compositions

If $f$ has period $T$, then $g \circ f$ also has period $T$.

Summary

ConceptFormula/Property
Composition$(g \circ f)(x) = g(f(x))$
Not commutative$g \circ f \neq f \circ g$ in general
Associative$(h \circ g) \circ f = h \circ (g \circ f)$
Inverse existsOnly for bijective functions
Inverse property$f^{-1}(f(x)) = x = f(f^{-1}(x))$
Composition inverse$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

Practice Problems

  1. If $f(x) = \frac{x}{x+1}$ and $g(x) = \frac{1}{1-x}$, find $(g \circ f)(x)$ and its domain.

  2. If $f(x) = x^3 + 3x - 1$ and $g = f^{-1}$, find $g(3)$. (Hint: If $f(a) = 3$, then $g(3) = a$)

  3. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x$ for all $x$.

  4. If $(f \circ f)(x) = 4x - 3$ and $f(0) = 1$, find $f(x)$.

Quick Check
If $f(x) = 2x$ and $g(x) = x + 3$, verify that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

What’s Next?

You’ve completed the Sets, Relations and Functions chapter! Key connections:

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