Composition of Functions and Invertible Functions

Composition and invertibility tie the chapter together and frequently appear as a 4–5 mark question in CBSE. The rule is mechanical once you respect the order of composition.

Composition

Given f : A → B and g : B → C, the composite g∘f : A → C is defined by (g∘f)(x) = g(f(x)). Apply f FIRST, then g.

• Composition is associative: h∘(g∘f) = (h∘g)∘f.
• It is generally NOT commutative: g∘f ≠ f∘g in most cases.

Invertible functions

f : A → B is invertible if there exists g : B → A with g∘f = I_A and f∘g = I_B (identity maps). Then g is the inverse, written f⁻¹.

• Key theorem: f is invertible ⟺ f is bijective (one-one AND onto).
• To find f⁻¹: set y = f(x), solve for x in terms of y, then swap to write f⁻¹(y).

Worked example

f(x) = 2x + 3. Put y = 2x + 3 ⇒ x = (y − 3)/2. So f⁻¹(x) = (x − 3)/2. Check: f(f⁻¹(x)) = 2·(x−3)/2 + 3 = x ✓.

Properties of inverses

• (f⁻¹)⁻¹ = f
• (g∘f)⁻¹ = f⁻¹∘g⁻¹ (order reverses — like putting on and taking off socks and shoes).

Exam Tricks & Tips

• 🎯 Always compute g(f(x)) inside-out: substitute f(x) into g.
• 🎯 Prove invertibility by proving bijectivity first, then construct f⁻¹ explicitly.
• 🎯 (g∘f)⁻¹ = f⁻¹∘g⁻¹ — the order FLIPS; a very common exam trap.
• 🎯 Domain of g∘f is the domain of f (with outputs landing in g's domain).
• 🎯 If f⁻¹ exists, its domain is f's range and its range is f's domain.
• ❌ Common mistake: assuming g∘f = f∘g. Composition is not commutative — always check the order.

Expected exam pattern

A 4–5 mark question: prove f is invertible (bijective) on a given set and find a formula for f⁻¹; sometimes verify g∘f and f∘g equal identity.

Quick recap

g∘f means g(f(x)), apply f first. Invertible ⟺ bijective. Find f⁻¹ by solving y = f(x) for x. Remember (g∘f)⁻¹ = f⁻¹∘g⁻¹.

Cover the answer, recall, then check. 12 cards on composition and inverse functions.

Q1. Define the composite g∘f.
A1. (g∘f)(x) = g(f(x)) — apply f first, then g. Needs f : A→B, g : B→C.

Q2. Is composition commutative?
A2. No — in general g∘f ≠ f∘g.

Q3. Is composition associative?
A3. Yes — h∘(g∘f) = (h∘g)∘f.

Q4. When is a function invertible?
A4. Exactly when it is bijective (one-one and onto).

Q5. How do you find f⁻¹?
A5. Set y = f(x), solve for x in terms of y, then write f⁻¹(y) = that expression.

Q6. Find f⁻¹ for f(x) = 2x + 3.
A6. f⁻¹(x) = (x − 3)/2.

Q7. What is (f⁻¹)⁻¹?
A7. f itself.

Q8. State the reversal law for inverse of a composite.
A8. (g∘f)⁻¹ = f⁻¹∘g⁻¹ — the order reverses.

Q9. What two identities characterise g = f⁻¹?
A9. g∘f = I_A and f∘g = I_B (identity on each set).

Q10. If f : A→B is invertible, what is the domain of f⁻¹?
A10. B (the codomain/range of f); its range is A.

Q11. If f(x)=x+1 and g(x)=x², find (g∘f)(2).
A11. g(f(2)) = g(3) = 9.

Q12. Same f, g: find (f∘g)(2) and compare.
A12. f(g(2)) = f(4) = 5 ≠ 9, confirming non-commutativity.

If one machine turns cloth into shirts and a second turns shirts into packed parcels, chaining them gives a single machine from cloth to parcels. That chaining is composition, and asking "can I run the machine backwards?" is the question of invertibility.

Definition: Given f : A → B and g : B → C, their composition g ∘ f : A → C is defined by (g ∘ f)(x) = g(f(x)) — apply f first, then g. The codomain of f must match the domain of g.

Order matters

Composition is not commutative: in general g ∘ f ≠ f ∘ g. Example: f(x) = x + 1, g(x) = x². Then (g ∘ f)(x) = (x+1)² but (f ∘ g)(x) = x² + 1 — different functions.

Two properties do hold:

• Associativity: h ∘ (g ∘ f) = (h ∘ g) ∘ f, so we can drop brackets.
• Preservation: the composition of two one-one functions is one-one; of two onto functions is onto; hence composition of two bijections is a bijection.

Invertible functions

Definition: f : A → B is invertible if there exists a function g : B → A with g ∘ f = I_A (identity on A) and f ∘ g = I_B (identity on B). Such g is unique, called the inverse f⁻¹.

Fundamental theorem: f is invertible if and only if f is a bijection (one-one and onto). One-one guarantees each output traces back to a single input; onto guarantees every codomain element has an input to trace back to.

How to find f⁻¹ (three steps)

1. Write y = f(x).
2. Solve for x in terms of y.
3. Swap symbols: f⁻¹(y) = (that expression); the inverse's domain is f's range.

A neat identity: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ — the inverse of a composition reverses the order (socks-then-shoes rule: to undo, take shoes off first).

Worked example

Question: Let f : ℝ → ℝ, f(x) = 3x − 4. Show f is invertible and find f⁻¹. Then verify (f⁻¹ ∘ f)(x) = x.

Solution:
Step 1 — Bijection: f(x₁) = f(x₂) ⟹ 3x₁ − 4 = 3x₂ − 4 ⟹ x₁ = x₂ (one-one). For any y, x = (y+4)/3 ∈ ℝ (onto). So f is a bijection ⟹ invertible.
Step 2 — Find inverse: y = 3x − 4 ⟹ x = (y + 4)/3 ⟹ f⁻¹(x) = (x + 4)/3.
Step 3 — Verify: (f⁻¹ ∘ f)(x) = f⁻¹(3x − 4) = ((3x − 4) + 4)/3 = 3x/3 = x. ✓
Conclusion: f⁻¹(x) = (x + 4)/3, and it correctly undoes f.

Exam application

Composition problems frequently ask for (f ∘ g)(x), its domain, or to find f when g and f ∘ g are given (solve backwards). Invertibility questions ask you to (i) restrict domain/codomain so a non-bijection becomes bijective, then (ii) write the inverse — e.g. f(x) = x² becomes invertible as f : [0,∞) → [0,∞) with f⁻¹(x) = √x. The graphs of f and f⁻¹ are mirror images across the line y = x.

Shortcut: To check two functions are inverses, just confirm f(g(x)) = x; you rarely need to derive the inverse from scratch.

Key formulas

• Composition: (g ∘ f)(x) = g(f(x)); domain of f feeds into g.
• Associativity: (h ∘ g) ∘ f = h ∘ (g ∘ f).
• Invertible function: f is invertible iff it is bijective; then f⁻¹ ∘ f = f ∘ f⁻¹ = I (identity).
• Inverse of composition: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹.
• Finding f⁻¹: set y = f(x), solve for x in terms of y.

Composition applies functions in sequence; only bijective functions have inverses that undo them.

Worked Example

Problem: Let f(x) = 2x + 3 and g(x) = x². Find (a) (f ∘ g)(x) and (g ∘ f)(x), and (b) the inverse of f.

Solution:
Step 1 — Compute (f ∘ g)(x) = f(g(x)). Substitute g(x) = x² into f:
(f ∘ g)(x) = f(x²) = 2x² + 3.

Step 2 — Compute (g ∘ f)(x) = g(f(x)). Substitute f(x) = 2x + 3 into g:
(g ∘ f)(x) = g(2x + 3) = (2x + 3)².
(Note f ∘ g ≠ g ∘ f in general.)

Step 3 — Find f⁻¹. Set y = f(x) = 2x + 3 and solve for x:
y = 2x + 3 ⟹ x = (y − 3)/2.

Step 4 — Write the inverse function (replace y by x):
f⁻¹(x) = (x − 3)/2.

Answer: (f ∘ g)(x) = 2x² + 3, (g ∘ f)(x) = (2x + 3)², and f⁻¹(x) = (x − 3)/2.