Find inverses algebraically, verify with composition, graph as a reflection over y = x — with worked examples.
A function where each output corresponds to exactly one input. Only one-to-one functions have inverses. Test: horizontal line test — every horizontal line hits the graph at most once.
Written as f⁻¹(x). This is NOT the same as 1/f(x). The inverse function undoes the original: if f(a) = b, then f⁻¹(b) = a.
f and g are inverses if and only if f(g(x)) = x AND g(f(x)) = x for all x in the respective domains.
The graph of f⁻¹ is the reflection of f over the line y = x. Swap all (x, y) coordinates: point (a, b) on f becomes (b, a) on f⁻¹.
The domain of f⁻¹ = range of f. The range of f⁻¹ = domain of f. This is a quick check for your answer.
When a function is not one-to-one, restrict the domain to make it one-to-one. Example: f(x) = x² is one-to-one on [0, ∞), giving f⁻¹(x) = √x.
Check one-to-one
Graph it or use the horizontal line test. If not one-to-one, state the domain restriction needed.
Replace f(x) with y
Write the equation as y = [formula]
Swap x and y
Write x = [formula with y]. This is the key step that finds the inverse.
Solve for y
Isolate y on one side using algebra.
Write as f⁻¹(x)
Replace y with f⁻¹(x). State the domain of f⁻¹ (= range of f).
Verify
Compose: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
f(x) = 3x − 7. Find f⁻¹(x).
y = 3x − 7
Swap: x = 3y − 7
x + 7 = 3y
y = (x + 7) / 3
f⁻¹(x) = (x + 7) / 3
Verify: f(f⁻¹(x)) = 3·(x+7)/3 − 7 = x + 7 − 7 = x ✓
f(x) = (2x + 1) / (x − 3). Find f⁻¹(x).
y = (2x + 1) / (x − 3)
Swap: x = (2y + 1) / (y − 3)
x(y − 3) = 2y + 1
xy − 3x = 2y + 1
xy − 2y = 3x + 1
y(x − 2) = 3x + 1
f⁻¹(x) = (3x + 1) / (x − 2)
Domain of f⁻¹: x ≠ 2 (also x = 3 in the range of f corresponds to asymptote in f⁻¹)
f(x) = x² + 2, x ≥ 0. Find f⁻¹(x).
Domain restricted to [0, ∞) → one-to-one
y = x² + 2
x = y² + 2 (after swap)
y² = x − 2
y = ±√(x − 2)
Since domain of f was x ≥ 0, range of f⁻¹ is y ≥ 0 → take positive root
f⁻¹(x) = √(x − 2), domain: x ≥ 2
| Function f(x) | Inverse f⁻¹(x) | Verification |
|---|---|---|
| b^x | log_b(x) | log_b(b^x) = x ✓ |
| e^x | ln(x) | ln(e^x) = x ✓ |
| log_b(x) | b^x | b^(log_b(x)) = x ✓ |
| sin(x), x ∈ [−π/2, π/2] | arcsin(x) | arcsin(sin x) = x on restricted domain ✓ |
| cos(x), x ∈ [0, π] | arccos(x) | arccos(cos x) = x on restricted domain ✓ |
| tan(x), x ∈ (−π/2, π/2) | arctan(x) | arctan(tan x) = x on restricted domain ✓ |
To find the inverse of y = f(x): (1) Replace f(x) with y; (2) Swap x and y — write x = f(y); (3) Solve for y; (4) Write the result as f⁻¹(x). Example: f(x) = 2x + 3 → x = 2y + 3 → y = (x−3)/2 → f⁻¹(x) = (x−3)/2.
Compose f and f⁻¹ in both orders and verify both equal x: f(f⁻¹(x)) = x AND f⁻¹(f(x)) = x. You must check both compositions. If either fails to simplify to x, they are not inverses.
A function has an inverse function (is one-to-one) if and only if every horizontal line intersects the graph at most once. If a horizontal line hits the graph more than once, the function is not one-to-one and does not have an inverse over its entire domain — you must restrict the domain first.
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