An inverse function, written f−1(x), is the function that undoes what f(x) does. If f doubles a number, f−1 halves it. If f adds 5, f−1 subtracts 5. Inverses are essential for solving equations and they have a beautiful graph property: they’re mirror images across the line y = x.
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What you need to know
f−1(x) reverses the effect of f(x): if f(2) = 5, then f−1(5) = 2
To find an inverse: swap x and y, then rearrange for y
Graph of y = f−1(x) is the reflection of y = f(x) in the line y = x
The domain and range swap: domain of f = range of f−1, and vice versa
An inverse exists only if the original function is one-to-one
f−1(x) is NOT the same as 1f(x)
What is an Inverse Function?
An inverse function undoes the original function. It takes the output and gives back the original input:
Function and Inverse — The “Undo” Pair
Input
x
f(x) →
← f−1(x)
Output
f(x)
Apply f then f−1 (or vice versa) → you’re back where you started.
Some easy examples:
If f(x) = 2x, then f−1(x) = x2 (doubling is undone by halving)
If g(x) = x + 10, then g−1(x) = x − 10 (adding 10 is undone by subtracting 10)
Solving Equations Using Inverses
If you know the inverse, you can solve f(x) = b in one move:
If f(x) = b, then x = f−1(b)
For example, if f(x) = 2x and you want to solve f(x) = 8, just compute x = f−1(8) = 82 = 4.
Watch the notation!f−1(x) is the inverse function — not the reciprocal. 1f(x) = [f(x)]−1 is something completely different. The “−1” sits as a superscript on the function name, not on the value.
How to Find an Inverse
The mechanical method is just two steps:
1
Swap
y = f(x) → x = f(y)
Replace every x with y, and vice versa.
2
Rearrange
x = f(y) → y = f−1(x)
Make y the subject. The result is f−1(x).
The Identity Function
When you apply f and then f−1 (or vice versa), you get back exactly what you started with. This is captured by the identity function, written id, which sends every value to itself:
id(x) = x and (f ∘ f−1)(x) = (f−1 ∘ f)(x) = x
So id(5) = 5, id(a) = a, and so on. Composing a function with its inverse always gives the identity.
Reflection in y = x
The graph of f−1(x) is the mirror image of f(x) reflected across the line y = x:
Mirror Image Across y = x
y = f(x) = 2x
y = f−1(x) = x/2
y = x (mirror line)
Every point (a, b) on y = f(x) corresponds to a point (b, a) on y = f−1(x) — the coordinates simply swap.
Useful trick: if f(x) crosses the line y = x at some point, then f−1(x) crosses it at the same point (because that point is its own reflection). So solutions to f(x) = x are also solutions to f(x) = f−1(x).
Domain and Range Swap
Inverses swap inputs and outputs — so the domain and range trade places:
Function f(x)
DomainD
RangeR
Inverse f−1(x)
DomainR
RangeD
So if f(x) = 2x has domain 1 ≤ x ≤ 3 (and range 2 ≤ f(x) ≤ 6), then f−1(x) has domain 2 ≤ x ≤ 6 and range 1 ≤ f−1(x) ≤ 3.
When Does an Inverse Exist?
For f−1(x) to be a function, the original f(x) must be one-to-one. Otherwise the “inverse” would be one-to-many, which isn’t allowed for functions.
For example, f(x) = x2 is many-to-one (both 2 and −2 map to 4) — so it doesn’t have an inverse on all of ℝ. But if you restrict the domain to x ≥ 0, the function becomes one-to-one and the inverse f−1(x) = √x works.
Worked Examples
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Example 1 — Linear inverse
Find the inverse of f(x) = 3x − 7.
Answer:
Step 1: write y = f(x), then swap x and y.y = 3x − 7x = 3y − 7Step 2: rearrange for y.x + 7 = 3yy = (x + 7)/3f⁻¹(x) = (x + 7)/3Check: f⁻¹(f(2)) = f⁻¹(−1) = 6/3 = 2 ✓
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Example 2 — Square root inverse
Find the inverse of f(x) = √(2x − 1), with domain x ≥ 1/2.
Answer:
Step 1: write y = f(x), then swap.y = √(2x − 1)x = √(2y − 1)Step 2: rearrange for y.x² = 2y − 12y = x² + 1y = (x² + 1)/2f⁻¹(x) = (x² + 1)/2, x ≥ 0The domain x ≥ 0 comes from the range of f (since √ outputs are always ≥ 0).
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Example 3 — Use inverse to solve an equation
Given f(x) = 2x + 3, solve f(x) = 11 using the inverse function.
Answer:
Step 1: find f⁻¹(x) by swap-and-rearrange.y = 2x + 3 → x = 2y + 3y = (x − 3)/2f⁻¹(x) = (x − 3)/2Step 2: use the rule f(x) = b ⟺ x = f⁻¹(b).x = f⁻¹(11) = (11 − 3)/2 = 8/2x = 4Check: f(4) = 2(4) + 3 = 11 ✓
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Example 4 — Domain and range swap
A function f(x) = (x + 3)/2 has domain 1 ≤ x ≤ 7. Find the inverse and state its domain and range.
Answer:
Step 1: find the inverse.y = (x + 3)/2 → x = (y + 3)/22x = y + 3 → y = 2x − 3f⁻¹(x) = 2x − 3Step 2: find the range of f (using the domain).f(1) = 4/2 = 2, f(7) = 10/2 = 5Range of f: 2 ≤ f(x) ≤ 5Step 3: swap domain and range for f⁻¹.Domain of f⁻¹: 2 ≤ x ≤ 5Range of f⁻¹: 1 ≤ f⁻¹(x) ≤ 7
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Example 5 — Rational function (full problem)
For the function f(x) = 2xx − 1, with x > 1:
(a) Find the inverse f−1(x).
Step 1: write y = f(x), then swap.y = 2x/(x − 1) → x = 2y/(y − 1)Step 2: rearrange for y.x(y − 1) = 2yxy − x = 2yxy − 2y = xy(x − 2) = xy = x/(x − 2)f⁻¹(x) = x/(x − 2)
(b) Find the domain of f−1(x).
Domain of f⁻¹ = range of f.Sketch f(x) = 2x/(x−1) for x > 1: as x → 1⁺, f(x) → ∞; as x → ∞, f(x) → 2.So for x > 1, f(x) > 2.Domain of f⁻¹: x > 2
(c) Find the value of k such that f(k) = 6.
Use the rule: f(k) = 6 ⟺ k = f⁻¹(6).k = f⁻¹(6) = 6/(6 − 2) = 6/4k = 3/2Check: f(3/2) = 3 / (1/2) = 6 ✓
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Tips
Always check by composing: if you’ve found the right inverse, then f(f−1(x)) should simplify to x. Quick sanity check.
Sketch f(x) to find its range when you need the domain of the inverse. The range of f = the domain of f−1.
Use the inverse to solve f(x) = b: just compute x = f−1(b). Faster than solving the original equation.
Reflection sketch shortcut: if you have f(x), you can sketch f−1(x) by flipping the graph across y = x — no algebra needed.
For square roots, remember the range of √(…) is always ≥ 0. This becomes a domain restriction on the inverse.
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Common mistakes
Confusing f−1(x) with 1f(x). The inverse function reverses f; the reciprocal flips the value upside down. Totally different things.
Forgetting to swap x and y before rearranging. Without the swap, you just rearrange f(x) for x — which doesn’t give you the inverse.
Forgetting the domain restriction on the inverse. The domain of f−1 = the range of f. Just writing x ∈ ℝ misses important detail.
Trying to find the inverse of a many-to-one function without restricting the domain. x2, sin x, etc., need a restricted domain first.
Sign errors when rearranging. Rational and surd inverses involve careful algebra — slow down and check each step.
Solving f(x) = b and forgetting to use the inverse. If the question gives you the inverse, use it — that’s why it’s there.
Final word: Inverse = swap and rearrange. Graphs reflect across y = x. Domain and range trade places. That’s the whole topic — practice the algebra and you’ll never miss these marks.
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