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Inverse function calculator

--- --- Methodology

The inverse of a function undoes it: if f turns 2 into 5, f⁻¹ turns 5 back into 2. Pick the form of f(x), enter its coefficients and the calculator shows the swap-and-solve steps, the inverse f⁻¹(x), the domain and range of both functions and the value of f⁻¹ at any y you choose.

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Inverse function f⁻¹(x) −1 + √(x + 4)
f(x)x² + 2x − 3
f⁻¹(x)−1 + √(x + 4)
Domain of f usedx ≥ −1
Range of fy ≥ −4
Domain of f⁻¹ (= range of f)x ≥ −4
Range of f⁻¹ (= domain of f)y ≥ −1
f⁻¹(5)2
Check: f(f⁻¹(5))5
Show the working, step by step

  1. Write y = f(x).

    y = x² + 2x − 3

  2. Complete the square to see the vertex.

    y = (x + 1)² − 4

  3. Restrict the domain so f is one-to-one: x ≥ −1.

    On this branch the vertex is at (−1, −4) and every y in the range comes from exactly one x.

  4. Swap x and y.

    x = (y + 1)² − 4

  5. Solve for y: isolate the square, then take the square root, keeping the + root because y ≥ −1.

    y = −1 + √(x + 4)

  6. State the result with its domain.

    f⁻¹(x) = −1 + √(x + 4), for x ≥ −4

  7. Evaluate at y = 5 and check by applying f.

    f⁻¹(5) = 2 f(2) = 5

Reading f⁻¹ as 1/f is the common slip: the inverse undoes f, the reciprocal divides 1 by it. The graphs of f and f⁻¹ mirror each other in the line y = x.

How to find an inverse: swap and solve

  1. Write the function as y = f(x).
  2. Swap x and y.
  3. Solve the new equation for y. That y is f⁻¹(x).
  4. State the domain of f⁻¹, which is the range of f, and check that f(f⁻¹(x)) = x.

An inverse exists only when f is one-to-one. If it is not, restrict the domain first, as below. The function calculator will find an inverse for any expression you type, and draws f, f⁻¹ and y = x together. This page works from the standard families, so it can show the domain restriction, the full domain and range, and each step.

The five forms and their inverses

f(x)f⁻¹(x)Domain of f⁻¹
ax + b(x − b)/aall real x
a/x + ba/(x − b)x ≠ b
a(x − h)² + k, x ≥ hh + √((x − k)/a)x ≥ k if a > 0, x ≤ k if a < 0
a·eˣ + bln((x − b)/a)(x − b)/a > 0
a·ln(x) + be^((x − b)/a)all real x

For x ≤ h the quadratic inverse takes the minus root: h − √((x − k)/a). In every row a ≠ 0, because a constant function gives one output for every input and cannot be undone.

Worked example: x² + 2x − 3 on x ≥ −1

The calculator starts with a = 1, b = 2, c = −3, the right branch and y = 5.

  1. Complete the square: x² + 2x − 3 = (x + 1)² − 4, so the vertex is (−1, −4).
  2. Restrict to x ≥ −1, the right of the vertex, so each output comes from one input. The range is y ≥ −4.
  3. Swap: x = (y + 1)² − 4.
  4. Solve: (y + 1)² = x + 4, so y + 1 = √(x + 4) (the + root, since y ≥ −1) and f⁻¹(x) = −1 + √(x + 4), with domain x ≥ −4.
  5. Evaluate at 5: f⁻¹(5) = −1 + √9 = 2. Check: f(2) = 4 + 4 − 3 = 5.

On the left branch (x ≤ −1) the inverse is −1 − √(x + 4), and f⁻¹(5) = −1 − 3 = −4. Check: f(−4) = 16 − 8 − 3 = 5. Both −4 and 2 map to 5, which is exactly why x² + 2x − 3 has no inverse until you choose a side.

Example: a reciprocal function

For f(x) = 2/x + 1, swap to x = 2/y + 1. Subtract 1: x − 1 = 2/y, so y = 2/(x − 1) and f⁻¹(x) = 2/(x − 1). The function f never outputs 1 and never takes the input 0; the inverse has exactly those two exclusions, swapped. At y = 3: f⁻¹(3) = 2/(3 − 1) = 1, and f(1) = 2/1 + 1 = 3.

Common mistakes

  • Reading f⁻¹ as 1/f. The inverse undoes f; the reciprocal divides 1 by it.
  • Forgetting the domain restriction. x² on all real numbers has no inverse; √x inverts only the x ≥ 0 branch.
  • Taking ± in the answer. An inverse is a function, so it gets one root. The restriction decides which.
  • Losing the exclusions. The inverse of a/x + b is not defined at x = b, even if the formula looks fine there.
  • Not checking. Apply f to your answer. If it does not return the input, an algebra step went wrong.

The inverse of a straight line is another straight line with the reciprocal slope; compare the two in the slope calculator. For the vertex form used above, the quadratic equation calculator shows the completed square.

Inverse function calculator: the worked example on this page, with its result and chart
Inverse function calculator: the worked example above, at a glance.

Common questions

How do you find the inverse of a function step by step?

Write y = f(x), swap x and y, then solve the new equation for y. For f(x) = 3x − 5: x = 3y − 5, so 3y = x + 5 and f⁻¹(x) = (x + 5)/3. Finally check by composing: f(f⁻¹(x)) = 3 · (x + 5)/3 − 5 = x.

How do you find the inverse of a quadratic function?

A quadratic fails the horizontal line test, so first restrict its domain to one side of the vertex. For f(x) = x² + 2x − 3 = (x + 1)² − 4 take x ≥ −1. Swap: x = (y + 1)² − 4, so y + 1 = √(x + 4) and f⁻¹(x) = −1 + √(x + 4). On the other side (x ≤ −1) the inverse is −1 − √(x + 4). The calculator asks for the branch and tells you if you leave it unrestricted.

Which functions have an inverse?

Only one-to-one functions: no two different inputs may give the same output. Graphically, every horizontal line crosses the graph at most once. Linear functions with a ≠ 0, eˣ and ln x pass; x² and sin x do not on their full domains, but do once restricted (x ≥ 0, or −90° ≤ x ≤ 90°).

What is the domain and range of an inverse function?

They swap. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. For f(x) = 2/x + 1 the domain is x ≠ 0 and the range is y ≠ 1, so f⁻¹(x) = 2/(x − 1) has domain x ≠ 1 and range y ≠ 0. That is why the calculator states both for every inverse.

Is the inverse function the same as 1/f(x)?

No. f⁻¹ undoes f, while 1/f is the reciprocal. For f(x) = 2x, f⁻¹(x) = x/2 but 1/f(x) = 1/(2x). The −1 in f⁻¹ is a label for the inverse, not an exponent.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.