MHT-CET Maths · Sets, Relations and Functions

Composite Functions — f∘g, Iteration and Functional Identities

(f∘g)(x) = f(g(x)): apply the inner function first, then the outer — evaluate numerically from the inside out, recover f from a given f(g(x)) by matching shapes, and prove a functional identity by simplifying the argument.

All Sets, Relations and Functions notes MHT-CET Maths strategy

Why this matters

21 PYQs at 10% HARD. The stems are a chain like f(g(g(f(1)))) evaluated step by step, f(f(x)) = x used to fix a parameter, and the log identity f(2x/(1 + x²)) = 2f(x) with its cubic twin. One HARD question asks for g∘g∘f as a formula and is answered by composing in two steps rather than one. Nothing here is more than substitution done in the right order.

Concept 1 of 4: Evaluate a Composite at a Point: Innermost First

f(g(g(f(1))))

is four separate evaluations. Start with

f(1)

, feed the number into

g

, feed that into

g

again, then into

f

. Never expand the formula.

Definition

$f(x) = x^2 + 1$ , $g(x) = \dfrac1x$ : $f(1) = 2 \to g(2) = \dfrac12 \to g\!\left(\dfrac12\right) = 2 \to f(2) = 5$ .
$f(x) = \dfrac{3x - 2}{5x + 3}$ : $f(1) = \dfrac18$ , $f\!\left(\dfrac18\right) = \dfrac{3/8 - 2}{5/8 + 3} = \dfrac{-13/8}{29/8} = -\dfrac{13}{29}$ .
$g(f(x))$ where $f$ is a trigonometric expression that simplifies to a CONSTANT: $\sin^2 x + \sin^2\!\left(x + \tfrac{\pi}{3}\right) + \cos x\cos\!\left(x + \tfrac{\pi}{3}\right) = \dfrac54$ for all $x$ , so $g(f(x)) = g\!\left(\dfrac54\right)$ , whatever $g$ is.
Order matters: $f \circ g \ne g \circ f$ in general; when they ARE equal ( $f(x) = \dfrac{3x + 4}{5x - 7}$ , $g(x) = \dfrac{7x + 4}{5x - 3}$ ), it is because each is the other's inverse and both composites equal $x$ .

Composition

(f \circ g)(x) = f(g(x)) \qquad \text{evaluate inside} \to \text{outside}

Worked example

f(x) = 2x - 1

and

g(x) = x^2

, find

f(g(f(2)))

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q126Easy

Example 1 · Sets, Relations and Functions · Composite Functions — f∘g, Iteration and Functional Identities

f(x) = x^2+1

and

g(x) = \frac{1}{x}

, then the value of

f(g(g(f(x))))

x = 1

Reading f(g(g(f(x)))) as (f∘g)² or as f²g²

It is a four-step chain, evaluated one function at a time from the inside. Option (A)

4

f(g(1))

-style short-cutting; the chain gives

5

Drill 3 more on evaluate a composite at a point: innermost first

Concept 2 of 4: Composite as a Formula: Substitute in Two Steps, Then Simplify

For

g \circ g \circ f

, first find

g(g(u))

in terms of

u

, then put

u = f(x)

. Composing a rational function with itself first keeps the algebra to one fraction.

Definition

$g(x) = \dfrac{x + 1}{x + 2}$ : $g(g(u)) = \dfrac{\frac{u + 1}{u + 2} + 1}{\frac{u + 1}{u + 2} + 2} = \dfrac{2u + 3}{3u + 5}$ . With $u = f(x) = \dfrac{x}{2 - x}$ : $2u + 3 = \dfrac{6 - x}{2 - x}$ and $3u + 5 = \dfrac{10 - 2x}{2 - x}$ , so $(g \circ g \circ f)(x) = \dfrac{6 - x}{10 - 2x}$ — the denominators cancel in the ratio, which is why the answer is clean.\n- $f(f(x))$ for $f(x) = \dfrac{\alpha x}{x + 1}$ : $\dfrac{\alpha^2 x}{(\alpha + 1)x + 1}$ . Equal to $x$ for all $x$ needs $\alpha + 1 = 0$ and $\alpha^2 = 1$ : $\alpha = -1$ .
$(f \circ g)(x)$ with $f(x) = 2x - 3$ , $g(x) = x^3 + 5$ : $2x^3 + 7$ . With $f(x) = 3x + 10$ , $g(x) = x^2 - 1$ : $3x^2 + 7$ .
Simplify the inner composite fully before substituting the outer; a nested fraction left unsimplified is where sign errors live.

Two-step composition

(g \circ g \circ f)(x) = g\big(g(u)\big)\Big|_{u = f(x)}

Worked example

f(x) = x + 1

and

g(x) = \dfrac{x}{x + 1}

, find

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q147Moderate

Example 2 · Sets, Relations and Functions · Composite Functions — f∘g, Iteration and Functional Identities

Let

f(x) = \frac{\alpha x}{x+1},\, x \neq -1

, then for

\alpha = \,?

f(f(x)) = x

Matching α² = 1 alone

\alpha = 1

satisfies

\alpha^2 = 1

but not

\alpha + 1 = 0

;

f(f(x)) = x

needs BOTH coefficient equations, so

\alpha = -1

, option (D).

Drill 1 more on composite as a formula: substitute in two steps, then simplify

Concept 3 of 4: Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x

f(g(x))

can be rewritten as an expression in

g(x)

alone, that expression IS

f

. If only a number is asked, pick the

x

that makes

g(x)

the wanted input.

Definition

$g(x) = 1 + \sqrt x$ , $f(g(x)) = 3 + 2\sqrt x + x = (1 + \sqrt x)^2 + 2 = g(x)^2 + 2$ , so $f(x) = x^2 + 2$ and $f(f(x)) = (x^2 + 2)^2 + 2 = x^4 + 4x^2 + 6$ .
$g(x) = x^2 + x - 1$ , $(g \circ f)(x) = 4x^2 - 10x + 5$ , find $f(2)$ : put $x = 2$ : $g(f(2)) = 16 - 20 + 5 = 1$ , so $f(2)^2 + f(2) - 2 = 0$ , $f(2) = 1$ or $-2$ ; the option list carries $1$ .
Look for a perfect square or a known expansion in the given composite; $3 + 2\sqrt x + x$ is $(1 + \sqrt x)^2 + 2$ , not a coincidence.

Shape matching

f(g(x)) = \Phi\big(g(x)\big) \ \Rightarrow\ f = \Phi

Worked example

g(x) = 2x + 1

and

f(g(x)) = 4x^2 + 4x + 3

, find

f(x)

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q106Moderate

Example 3 · Sets, Relations and Functions · Composite Functions — f∘g, Iteration and Functional Identities

g(x)=1+\sqrt{x}

and

f(g(x))=3+2\sqrt{x}+x

, then

f(f(x))

Solving for f(x) when only f(2) is asked

Finding

f

as a formula from

(g \circ f)(x) = 4x^2 - 10x + 5

needs a square root of a quadratic. One substitution,

x = 2

, gives a quadratic in the number

f(2)

Drill 1 more on recover f from f(g(x)): match the shape, or evaluate at the right x

Concept 4 of 4: Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin

For

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

, substituting a rational expression and factoring

1 \mp u

turns the argument into a power of

\dfrac{1 - x}{1 + x}

; the log pulls the power out front.

Definition

$u = \dfrac{2x}{1 + x^2}$ : $1 - u = \dfrac{(1 - x)^2}{1 + x^2}$ , $1 + u = \dfrac{(1 + x)^2}{1 + x^2}$ , so $\dfrac{1 - u}{1 + u} = \left(\dfrac{1 - x}{1 + x}\right)^2$ and $f(u) = 2f(x)$ .
$f(x) = \log\dfrac{1 + x}{1 - x}$ , $g(x) = \dfrac{3x + x^3}{1 + 3x^2}$ : $1 + g = \dfrac{(1 + x)^3}{1 + 3x^2}$ , $1 - g = \dfrac{(1 - x)^3}{1 + 3x^2}$ , so $(f \circ g)(x) = 3f(x)$ .
The pattern: $\dfrac{2x}{1 + x^2}$ is the $\tanh$ -double-angle shape, $\dfrac{3x + x^3}{1 + 3x^2}$ the triple; the log of the ratio scales by $2$ and $3$ respectively.
Compute $1 - u$ and $1 + u$ SEPARATELY over the common denominator; the denominators cancel in the ratio, which is why the answer is clean.

The two identities

f(x) = \log\frac{1 - x}{1 + x}:\quad f\!\left(\frac{2x}{1 + x^2}\right) = 2f(x),\qquad f\!\left(\frac{3x + x^3}{1 + 3x^2}\right) = 3f(x)

Worked example

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

, show that

f(a) + f(b) = f\!\left(\dfrac{a + b}{1 + ab}\right)

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q133Moderate

Example 4 · Sets, Relations and Functions · Composite Functions — f∘g, Iteration and Functional Identities

f(x)=\log_e\!\frac{1-x}{1+x},\,|x|<1

, then

f\!\left(\frac{2x}{1+x^2}\right)

is equal to

Reading the sign of the ratio backwards

\dfrac{1 - u}{1 + u}

with

u = \dfrac{2x}{1 + x^2}

\left(\dfrac{1 - x}{1 + x}\right)^2

, positive power. Option (B)

-2f(x)

is the inverted ratio.

Drill 1 more on functional identities: f(2x/(1 + x²)) = 2f(x) and the cubic twin

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 (4)

Evaluate a Composite at a Point: Innermost First
Composition
$(f \circ g)(x) = f(g(x)) \qquad \text{evaluate inside} \to \text{outside}$
Composite as a Formula: Substitute in Two Steps, Then Simplify
Two-step composition
$(g \circ g \circ f)(x) = g\big(g(u)\big)\Big|_{u = f(x)}$
Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x
Shape matching
$f(g(x)) = \Phi\big(g(x)\big) \ \Rightarrow\ f = \Phi$
Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin
The two identities
$f(x) = \log\frac{1 - x}{1 + x}:\quad f\!\left(\frac{2x}{1 + x^2}\right) = 2f(x),\qquad f\!\left(\frac{3x + x^3}{1 + 3x^2}\right) = 3f(x)$

Watch out for (4)

Reading f(g(g(f(x)))) as (f∘g)² or as f²g²
→ Evaluate a Composite at a Point: Innermost First
Matching α² = 1 alone
→ Composite as a Formula: Substitute in Two Steps, Then Simplify
Solving for f(x) when only f(2) is asked
→ Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x
Reading the sign of the ratio backwards
→ Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin

Test yourself on Sets, Relations and Functions

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