Composition of functions and invertible functions

What is the composition of two functions?

In plain words: to find gof(x), feed x to f, then feed the answer f(x) to g. This needs every output of f to be an input that g accepts, which is why the codomain B of f is the domain of g. gof goes straight from A to C, skipping B. More generally it is enough that the range of f lies inside the domain of g; the two named sets need not be identical.

Domains of real composites (JEE). Beyond the NCERT text, for real functions given by formulas gof(x) is defined only where f(x) is defined and f(x) lies in the domain of g. For f(x) = x − 4 and g(x) = √x, gof(x) = √(x − 4) needs x ≥ 4, while fog(x) = √x − 4 needs only x ≥ 0.

gof and fog are usually different

NCERT Example 16 (page 12): f(x) = cos x and g(x) = 3x² on R.

• gof(x) = g(cos x) = 3 cos² x;
• fog(x) = f(3x²) = cos 3x².

At x = 0 the first is 3 and the second is cos 0 = 1, so gof ≠ fog. One value is enough to show that two functions differ.

Sometimes only one order makes sense. In NCERT Example 15, g ends in {7, 11, 15}, and none of these is in the domain {2, 3, 4, 5} of f, so fog is not defined at all.

Beyond the NCERT text, three facts reasoned from the definitions:

• composition is associative: ho(gof) and (hog)of both send x to h(g(f(x)));
• if f and g are both one-one, so is gof; if both are onto, so is gof;
• the converses hold only halfway: if gof is one-one then f is one-one, and if gof is onto then g is onto, but g need not be one-one nor f onto (f : {1} → {1, 2}, f(1) = 1, and g : {1, 2} → {1} give a bijection gof).

Invertible functions

NCERT Definition 9 (page 12): a function f : X → Y is invertible if there is a function g : Y → X such that gof = I_X and fog = I_Y, where I_X(x) = x is the identity function of X. This g is the inverse of f, written f⁻¹. So f⁻¹ undoes f, and f undoes f⁻¹:

Thus, if f is invertible, then f must be one-one and onto and conversely, if f is one-one and onto, then f must be invertible.

NCERT Mathematics Part I (Class 12), Section 1.4, page 12

NCERT states this without proof. The reason, reasoned from Definition 9:

• Invertible ⇒ one-one and onto. If f(x_₁) = f(x_₂), apply g to both sides: x_₁ = g(f(x_₁)) = g(f(x_₂)) = x_₂. And any y ∈ Y equals f(g(y)), so it is an image.
• One-one and onto ⇒ invertible. Onto gives each y at least one x with f(x) = y, and one-one makes that x unique. Let g(y) be this x. Then f(g(y)) = y by the choice of g(y), and g(f(x)) = x because x is the only element with image f(x). So gof = I_X and fog = I_Y.

So the one-one and onto tests decide invertibility, often without finding the inverse, as NCERT notes on page 12.

Finding the inverse. Check that f is one-one and onto, write y = f(x), solve for x, and name the result f⁻¹(y). Then check that f⁻¹(f(x)) = x and f(f⁻¹(y)) = y.

The graph of the inverse. (a, b) is on the graph of f exactly when (b, a) is on the graph of f⁻¹, so the two graphs are mirror images in the line y = x (NCERT states this in Chapter 2, pages 19–20).

Two more facts (JEE). Beyond the NCERT text, and following from the definition: (f⁻¹)⁻¹ = f, and if f and g are invertible, then (gof)⁻¹ = f⁻¹og⁻¹: to undo “f, then g”, undo g first.

Solved examples

Common mistakes to avoid

Quick revision summary

• Composition: gof(x) = g(f(x)), defined when the outputs of f are inputs of g; f acts first.
• Order matters: in general gof ≠ fog, and one of them may not even be defined.
• f : X → Y is invertible when some g : Y → X has gof = I_X and fog = I_Y; that g is f⁻¹.
• f is invertible exactly when it is one-one and onto. To find f⁻¹, solve y = f(x) for x.

Questions students ask

Can gof and fog ever be equal?

Yes, gof and fog are equal for some pairs of functions f, g : X → X, even though they differ in general. Two translations always commute, as f(x) = x + 4 and g(x) = x − 6 give x − 2 in both orders, and so do a bijection f : X → X and its inverse, since f⁻¹of and fof⁻¹ are both I_X. For linear functions the condition can be solved: with f(x) = 3x + 1 and g(x) = 7x + k, fog(x) = 21x + 3k + 1 and gof(x) = 21x + 7 + k, which agree for every x only when k = 3. Commuting does not make f and g inverses: x + 4 and x − 6 commute, yet neither undoes the other.

Is the inverse of a function the same as its inverse relation?

Every function f : X → Y has an inverse relation, its pairs reversed as for any relation, but that relation is the inverse function f⁻¹ only when f is one-one and onto. Reversing f = {(1, 10), (2, 10), (3, 20)} from {1, 2, 3} to {10, 20, 30} gives {(10, 1), (10, 2), (20, 3)}, which is not a function from {10, 20, 30} to {1, 2, 3}: 10 has two images because f is not one-one, and 30 has none because f is not onto. When f is one-one and onto, each y in Y is paired with exactly one x, and the reversed relation is f⁻¹, with f⁻¹(y) = x exactly when f(x) = y.

How do you find the inverse of a quadratic function?

A quadratic function is never one-one on all of R, as its graph is symmetric about the vertical line through its vertex, so a usual way to make it invertible is to restrict its domain to an interval on one side of the vertex and its codomain to the resulting range. More generally, any restriction that keeps at most one input for each output is one-one. For f(x) = x² − 6x + 10 = (x − 3)² + 1, take f : [3, ∞) → [1, ∞); solving y = (x − 3)² + 1 with x ≥ 3 gives f⁻¹(y) = 3 + √(y − 1), so f⁻¹(2) = 4. For f : (−∞, 3] → [1, ∞) the inverse is 3 − √(y − 1). Inverse trigonometric functions are made the same way, by restricting the domain.

What is the difference between an invertible function and an inverse function?

“Invertible” is a property of a function f : X → Y, that some g : Y → X has gof = I_X and fog = I_Y, while “the inverse” is that function g itself, written f⁻¹; only an invertible function has an inverse. For f(x) = 7 − 2x on R, f is invertible because it is one-one and onto, and its inverse is f⁻¹(y) = (7 − y)/2, so f(5) = −3 and f⁻¹(−3) = 5. The inverse is invertible too, with inverse f, and it is unique: if g and h both have this property, then g = go(foh) = (gof)oh = I_Xoh = h.

Are composition of functions and inverse functions in the JEE Main 2027 syllabus?

Yes, composition of functions and inverse functions are both in the JEE Main 2027 syllabus: Unit 1, Sets, Relations and Functions, ends with “one-one, into and onto functions, the composition of functions”, and Unit 7, Limit, Continuity and Differentiability, lists “inverse functions” among its real-valued functions. Unit 7 also covers the differentiation of “composite and implicit functions” and of inverse trigonometric functions, which build on gof and f⁻¹. Both ideas are taught in NCERT Mathematics Part I (Class 12), Chapter 1, Section 1.4 (pages 12–13).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. f(x) = x + 2 and g(x) = 3x on R. Find gof(x), fog(x) and gof(1).

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   gof(x) = 3(x + 2) = 3x + 6 and fog(x) = 3x + 2. gof(1) = 9, while fog(1) = 5.

2. Q2. f : {1, 2, 3} → {5, 6, 7} has f(1) = 6, f(2) = 7, f(3) = 5, and g : {5, 6, 7} → {0, 1} has g(5) = 0, g(6) = 1, g(7) = 1. Find gof. Is fog defined?

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   gof(1) = g(6) = 1, gof(2) = g(7) = 1 and gof(3) = g(5) = 0, so gof = {(1, 1), (2, 1), (3, 0)}. fog is not defined on the whole domain {5, 6, 7} of g: g(5) = 0 is not in the domain of f. A restricted composite on {6, 7} is possible, with both values equal to f(1) = 6.

3. Q3. Show that f : R → R, f(x) = 5x + 2, is invertible and find f⁻¹.

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   5x_₁ + 2 = 5x_₂ + 2 gives x_₁ = x_₂; and y = 5x + 2 gives x = (y − 2)/5 for every real y. So f is one-one and onto, and f⁻¹(y) = (y − 2)/5. Check: f((y − 2)/5) = y − 2 + 2 = y.

4. Q4. Is f : N → N, f(x) = x + 1, invertible? What if the codomain is {2, 3, 4, …}?

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   Into N it is one-one but not onto (1 is never x + 1), so it is not invertible. Into {2, 3, 4, …} it is also onto, and f⁻¹(y) = y − 1.