Function Composition & Inverse Functions: JEE Main

Function composition and inverse functions appear in JEE Main in the Functions chapter and also as setup tools in Calculus (chain rule, inverse trigonometry, and implicit differentiation). Students who understand these concepts structurally — not just the formulas — solve derivative of inverse function questions and composite domain questions much faster. This guide builds that structural understanding.

Function Composition: (f ∘ g)(x) = f(g(x))

Composition: apply g first, then f. The domain of f ∘ g: the set of x for which (1) x is in the domain of g, and (2) g(x) is in the domain of f. This two-condition domain restriction is the most tested aspect of composition in JEE. Example: f(x) = √x (domain: x ≥ 0), g(x) = x − 3. (f ∘ g)(x) = √(x−3). Domain: need x − 3 ≥ 0 → x ≥ 3. The range of g must overlap with the domain of f: g(x) ≥ 0 → x ≥ 3. (g ∘ f)(x) = √x − 3. Domain: x ≥ 0 (domain of f, which is applied first). Output: √x − 3; need √x − 3 to be in domain of g (which is all reals, no restriction) → domain = x ≥ 0.

Important: f ∘ g ≠ g ∘ f in general (composition is not commutative). f ∘ (g ∘ h) = (f ∘ g) ∘ h (composition is associative). Identity function I(x) = x: f ∘ I = I ∘ f = f. JEE tests whether f ∘ g = g ∘ f for specific functions — usually no. Exceptions: f and g are both linear with same slope, or f and g are inverses of each other. Take a free functions mock. See our function types guide.

Inverse Functions: When and How

A function f has an inverse f⁻¹ if and only if f is bijective (both injective and surjective — one-to-one and onto). Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f. Finding f⁻¹: write y = f(x), solve for x in terms of y → x = f⁻¹(y). Property: (f ∘ f⁻¹)(x) = x and (f⁻¹ ∘ f)(x) = x. Derivative relationship: if f and f⁻¹ are differentiable, then (f⁻¹)'(y) = 1/f'(f⁻¹(y)). JEE uses this to find the derivative of inverse trigonometric functions: d/dx[arcsin x] = 1/√(1−x²) (derived from the derivative relationship with sinx). Graph of f⁻¹ is the reflection of the graph of f in the line y = x. JEE asks for f⁻¹(a) when the graph of f passes through (a, b): the answer is b (since f(a) = b → f⁻¹(b) = a, so f⁻¹ of the output b is a — careful with which way the question is asked).

Inverse of a Composition

(f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹ (reverse order — "last in, first out"). This is analogous to putting on shoes and socks: to remove, take off shoes first, then socks. JEE tests this in the form: "if h = f ∘ g, find h⁻¹." Answer: h⁻¹ = g⁻¹ ∘ f⁻¹. Verification: (f ∘ g) ∘ (g⁻¹ ∘ f⁻¹) = f ∘ (g ∘ g⁻¹) ∘ f⁻¹ = f ∘ I ∘ f⁻¹ = f ∘ f⁻¹ = I ✓. Restriction: if f is not bijective over its full natural domain but is bijective over a restricted domain, define the inverse on that restricted domain. Example: f(x) = x² is not injective over ℝ (f(2) = f(−2) = 4), but it is injective over [0, ∞), giving f⁻¹(y) = √y. For inverse trigonometric functions (which are the most tested inverses in JEE), the principal value range is the restricted domain: arcsin: [−π/2, π/2]; arccos: [0, π]; arctan: (−π/2, π/2). For the inverse trigonometry chapter see our inverse trigonometry guide and for domain-range problems see our sets, relations and functions guide.