NDA Maths · Functions

Composition and Inverse of Functions

Chain functions together (f∘g) or run one backwards (f⁻¹) — the chapter's richest source of HARD questions.

Why this matters

Twenty-eight PYQs and the chapter's difficulty hot-spot — 7 of the chapter's HARD questions live here. The staples: evaluate f at a point, compose two functions (and notice f∘g ≠ g∘f), find the constant that makes two linear functions commute (a near-annual question), and invert a rational/exponential function. Order and bijectivity are where marks are won and lost.

Test yourself — a quick 16-question Functions recall quiz, instant score.

Concept 1 of 4: Evaluating and combining functions

To evaluate

f

at something, substitute it everywhere

x

appears — the input can be a number, another variable, or an expression like

\sin\theta

. Sums and products of functions are done pointwise:

(f+g)(x)=f(x)+g(x)

(fg)(x)=f(x)\,g(x)

Definition

Value: $f(a)$ replaces $x$ by $a$ throughout the rule.
Substitution: $f(\sin\theta)$ replaces $x$ by $\sin\theta$ (use identities to simplify).
Pointwise sum/product: $(f\pm g)(x)=f(x)\pm g(x)$ , $(fg)(x)=f(x)g(x)$ , $\left(\tfrac fg\right)(x)=\tfrac{f(x)}{g(x)}$ where $g(x)\neq0$ .

Worked example

f(x)=x+1

and

g(x)=x^2

, find

(fg)(2)

The same idea in a real exam question:

NDA · Apr 2023 · Q96Easy

Example 1 · Functions · Composition and Inverse of Functions

f(x)=x^{2}+2

and

g(x)=2x-3

, then what is

(fg)(1)

equal to?

$(fg)$ is a product, $(f\circ g)$ is a composition

NDA writes the product as

(fg)(x)=f(x)g(x)

and the composition as

(f\circ g)(x)=f(g(x))

. They are completely different operations — read the symbol carefully before computing.

Drill 5 more on evaluating and combining functions

Concept 2 of 4: Composition of functions

Composition feeds one function's output into the next:

(f\circ g)(x)=f(g(x))

means 'do

g

first, then

f

'. Work inside-out, and remember the order matters —

f\circ g

and

g\circ f

are usually different.

Definition

$(f\circ g)(x)=f(g(x))$ ; the range of $g$ must sit inside the domain of $f$ . Composition is associative $(f\circ(g\circ h)=(f\circ g)\circ h)$ but not commutative $(f\circ g\neq g\circ f$ in general $)$ . Iterating $f\circ f\circ\cdots$ just repeats the substitution.

Composition and its inverse

(f\circ g)(x)=f(g(x))\qquad (f\circ g)^{-1}=g^{-1}\circ f^{-1}

Worked example

With

f(x)=2x+1

and

g(x)=x^2

, find

(f\circ g)(x)

and

(g\circ f)(x)

Practice this conceptself-check

The same idea in a real exam question:

NDA · Apr 2022 · Q68Easy

Example 2 · Functions · Composition and Inverse of Functions

f(x)=4x+3

, then what is

f\circ f\circ f(-1)

equal to?

Work inside-out, and keep the order

(f\circ g)(x)=f(g(x))

means

g

acts first. Students often apply

f

first or read

f\circ g

as a product. For iterated composition, peel one layer at a time — don't try to do all of

f\circ f\circ f

in one leap.

Drill 15 more on composition of functions

Concept 3 of 4: When do two linear functions commute?

A near-annual NDA question: given two linear functions, find the constant that makes

f\circ g=g\circ f

. Both compositions are linear, so just expand both, match coefficients, and solve.

Definition

For $f(x)=ax+b$ and $g(x)=cx+d$ : $f(g(x))=acx+ad+b$ and $g(f(x))=acx+bc+d$ . The $x$ -coefficients always match (both $ac$ ), so $f\circ g=g\circ f$ reduces to the constant condition $ad+b=bc+d$ , i.e. $b(c-1)=d(a-1)$ .

Commuting condition for linear f, g

f\circ g=g\circ f\iff b(c-1)=d(a-1)\ \ [f=ax+b,\ g=cx+d]

Worked example

f(x)=2x+1

and

g(x)=3x+d

satisfy

f\circ g=g\circ f

, find

d

The same idea in a real exam question:

NDA · Sep 2022 · Q72Moderate

Example 3 · Functions · Composition and Inverse of Functions

f(x) = 4x+1

and

g(x) = kx+2

such that

fog(x) = gof(x)

, then what is the value of k?

Drill 3 more on when do two linear functions commute?

Concept 4 of 4: Inverse of a function

The inverse

f^{-1}

undoes

f

: if

f

sends

a\mapsto b

, then

f^{-1}

sends

b\mapsto a

. It exists only when

f

is a bijection, and its graph is the mirror image of

f

across the line

y=x

Definition

To find $f^{-1}$ : write $y=f(x)$ , solve for $x$ in terms of $y$ , then swap names. Domain and range swap: $\text{dom}(f^{-1})=\text{range}(f)$ . Properties: $f^{-1}\circ f=\text{id}$ , and the graph of $f^{-1}$ is the reflection of $f$ in $y=x$ . Note $f^{-1}(x)\neq\dfrac{1}{f(x)}$ .

Self-inverse (involution): if $f(f(x))=x$ for all $x$ then $f^{-1}=f$ , and such an $f$ is automatically one-one AND onto (it has an inverse, namely itself). Examples: $f(x)=1-x$ , $f(x)=\tfrac1x$ , $f(x)=\tfrac{x+1}{x-1}$ , $f(x)=-x$ .
Möbius (linear-fractional) inverse: $y=\dfrac{ax+b}{cx+d}\Rightarrow x=\dfrac{dy-b}{a-cy}$ , so $f^{-1}(x)=\dfrac{dx-b}{-cx+a}$ — swap $a\leftrightarrow d$ and negate $b,c$ . Its domain excludes $x=\tfrac ac$ , which is exactly the value the original $f$ never reaches (its range excludes $\tfrac ac$ ); $f$ is self-inverse iff $a+d=0$ .

Inverse of a linear function

f(x)=ax+b\ (a\neq0)\ \Rightarrow\ f^{-1}(x)=\dfrac{x-b}{a}

Worked example

Find the inverse of

f(x)=\dfrac{3x+2}{x-1}

x\neq1

Practice this conceptself-check

The same idea in a real exam question:

NDA · Apr 2019 · Q86Moderate

Example 4 · Functions · Composition and Inverse of Functions

f(x)=\dfrac{x-2}{x+2}

x\neq-2

, then what is

f^{-1}(x)

equal to?

Inverse needs a bijection — and $f^{-1}\neq1/f$

Only one-one onto functions have an inverse;

x^2

\mathbb{R}

has none (not one-one). And

f^{-1}

is the undo function, not the reciprocal —

f^{-1}(x)

is generally nothing like

\dfrac{1}{f(x)}

. When finding it, remember the domain of

f^{-1}

is the range of

f

Drill 4 more on inverse of a function

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Composition of functions
Composition and its inverse
$(f\circ g)(x)=f(g(x))\qquad (f\circ g)^{-1}=g^{-1}\circ f^{-1}$
When do two linear functions commute?
Commuting condition for linear f, g
$f\circ g=g\circ f\iff b(c-1)=d(a-1)\ \ [f=ax+b,\ g=cx+d]$
Inverse of a function
Inverse of a linear function
$f(x)=ax+b\ (a\neq0)\ \Rightarrow\ f^{-1}(x)=\dfrac{x-b}{a}$

Watch out for (3)

$(fg)$ is a product, $(f\circ g)$ is a composition
→ Evaluating and combining functions
Work inside-out, and keep the order
→ Composition of functions
Inverse needs a bijection — and $f^{-1}\neq1/f$
→ Inverse of a function

Test yourself on Functions

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