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Finding an inverse function is a short algebra exercise until the function is not one-one, or the square root comes with a sign you have to choose. Then students lose marks on the domain, not the algebra. This post covers when an inverse exists, a reliable method, two worked examples and how to sketch the inverse from the original graph.
When does a function have an inverse?
Section 1.4 of the NCERT Class 12 Mathematics chapter on relations and functions defines f : X → Y as invertible if there is a g with g(f(x)) = x and f(g(y)) = y, and notes that f is invertible exactly when it is one-one and onto. So before finding an inverse, check two things:
One-one: no two inputs give the same output. On a graph, every horizontal line meets the curve at most once.
Onto: the codomain is the range. If it is not, shrink the codomain to the range.
The JEE (Main) 2026 syllabus lists "inverse functions" and "graphs of simple functions". The CBSE Mathematics curriculum for 2026-27 lists one-one and onto functions in Class XII but does not name invertible functions, though the NCERT section is still in the book. So this is mainly a JEE topic.
How do you find the inverse?
Write y = f(x) and state the domain and range of f.
Solve for x in terms of y. If you take a square root, use the domain to choose the sign.
Swap x and y to write f⁻¹(x). Its domain is the range of f.
Check: f⁻¹(f(a)) = a for one value of a.
Worked example 1: a rational function
Find the inverse of f(x) = (2x + 3)/(x − 1), x ≠ 1.
Set y = (2x + 3)/(x − 1). Then y(x − 1) = 2x + 3, so xy − 2x = y + 3, and x(y − 2) = y + 3. So x = (y + 3)/(y − 2), which needs y ≠ 2. The range of f is all reals except 2.
f⁻¹(x) = (x + 3)/(x − 2), for x ≠ 2.
Check: f(2) = 7/1 = 7, and f⁻¹(7) = 10/5 = 2.
Notice the swap: f has a vertical asymptote x = 1 and a horizontal asymptote y = 2. Its inverse has a vertical asymptote x = 2 and a horizontal asymptote y = 1.
Worked example 2: restricting the domain
f(x) = x² − 4x + 5 is not one-one on all reals: f(1) = f(3) = 2. Completing the square, f(x) = (x − 2)² + 1, so it is one-one on x ≥ 2, with range y ≥ 1.
Set y = (x − 2)² + 1. Then (x − 2)² = y − 1, so x − 2 = ±√(y − 1). Since x ≥ 2, take the positive root: x = 2 + √(y − 1).
f⁻¹(x) = 2 + √(x − 1), for x ≥ 1.
Check: f(5) = 25 − 20 + 5 = 10, and f⁻¹(10) = 2 + 3 = 5.
How do you sketch the inverse from the graph?
The point (a, b) is on f exactly when (b, a) is on f⁻¹. So the graph of f⁻¹ is the reflection of the graph of f in the line y = x. In practice:
Reflect a few key points: intercepts, the vertex, a point near each asymptote.
Swap the asymptotes: vertical becomes horizontal and the other way round.
Swap domain and range on your axes.
Where do f and f⁻¹ meet? If f is increasing, any meeting point lies on y = x, so solve f(x) = x. For example 2, (x − 2)² + 1 = x gives x² − 5x + 5 = 0, so x = (5 ± √5)/2. Only (5 + √5)/2 ≈ 3.62 satisfies x ≥ 2, so the curves meet once, at about (3.62, 3.62). For a decreasing function this shortcut can fail: f(x) = −x³ and its inverse f⁻¹(x) = −∛x both pass through (1, −1), which is not on y = x.
Which mistakes cost marks?
Writing f⁻¹(x) = 1/f(x). The inverse undoes f; the reciprocal does not.
Skipping the one-one check, then finding an "inverse" for x² on all reals.
Choosing the wrong sign of the square root, or both signs.
Forgetting to state the inverse's domain.
These are not just algebra slips. Paoletti and colleagues (2018) interviewed 25 pre-service teachers about inverse functions in algebraic and graphical forms, starting from earlier research arguing that teachers and students often do not develop connected meanings for function inverse. Linking the "swap x and y" algebra to the reflection picture is the connection to build. For more on one-one and onto checks, see our post on one-one and onto functions, and for shifting and stretching graphs, graph transformations for JEE.
If a definition here feels shaky, Fermi's Concept Library covers every concept in physics, chemistry, biology, mathematics and computing.
A practice routine
Find the inverses of three rational functions of the form (ax + b)/(cx + d), and check each with a value.
Take two quadratics, restrict each to the right of its vertex, and find the inverse with the correct root sign.
Sketch each function and its inverse on the same axes with y = x drawn in.
Mix in one exponential and log pair, since they are inverses of each other.
Now try f(x) = (3x − 1)/(x + 2): find f⁻¹, then check that f⁻¹(f(1)) = 1.
