1
Only a one-to-one function has an inverse.
One-to-one means no two inputs share an output.
f(x) = 3x − 6
one-to-one ✓
every output used once
f(x) = x²
not one-to-one ✗
f(3) and f(−3) are both 9
f(x) = x², x ≥ 0
one-to-one ✓
restricting the domain rescues it
The horizontal line test decides this, just as the vertical one decides whether
something is a function at all. If a horizontal line meets the graph twice, two inputs
share an output and no inverse exists.
2
Write y, swap x and y, solve for y.
y = 2x + 5
rename f(x) as y
x = 2y + 5
swap the letters
x − 5 = 2y
undo the +5 first
f⁻¹(x) = x − 52
undo the ×2 second
The undoing happens in reverse order: the function doubled and then added, so the
inverse subtracts and then halves. Socks before shoes; shoes off before socks.
3
Compose the two to prove it.
f(x) = 2x + 5, f⁻¹(x) = x − 52
the claim
f(f⁻¹(x)) = 2(x − 52) + 5
feed one into the other
= x − 5 + 5 = x ✓
back where we started
Both orders must give x. If f(f⁻¹(x)) = x but f⁻¹(f(x)) does not, the domain has been
restricted somewhere and the inverse only works on part of the function.
4
Domain and range trade places.
f(x) = √x
domain x ≥ 0, range y ≥ 0
starts at the origin
f⁻¹(x) = x²
domain x ≥ 0, range y ≥ 0
the restriction is inherited
graphs
mirror images
reflected in the line y = x
Swapping x and y in the algebra is exactly what reflecting in y = x does to the
picture. The two facts are the same fact.
5
Read an inverse value without finding the formula.
If f(a) = b, then f⁻¹(b) = a. The point (a, b) on f is the point (b, a) on f⁻¹.
If f(6) = −1, what is f⁻¹(−1)?
f sends 6 to −1
the given fact
f⁻¹ sends −1 back to 6
the inverse reverses it
f⁻¹(−1) = 6
no formula needed
If f(x) = x³ + 4, what is f⁻¹(12)?
f⁻¹(12) is the input with f(input) = 12
ask the question backwards
x³ + 4 = 12
set the rule equal to 12
x³ = 8, so x = 2
the cube root of 8
f(2) = 8 + 4 = 12 ✓
check
f⁻¹(12) is not 1f(12)
Working out f(12) = 1,732 and turning it over gives 11,732 — the reciprocal of an output, which has nothing to do with the inverse. The inverse asks which input made 12.
6
Variations: dividing last, a negative x, a function that undoes itself.
Dividing last: f(x) = x − 45
5x = y − 4
undo the ÷5 first
f⁻¹(x) = 5x + 4
then undo the −4
Check: f(9) = 1 and f⁻¹(1) = 9 ✓. f divided last, so the inverse multiplies first.
A negative x: f(x) = 7 − 2x
2y = 7 − x
move the y-term across
f⁻¹(x) = 7 − x2
divide by 2
Check: f(1) = 5 and f⁻¹(5) = 22 = 1 ✓. Moving −2y to the other side keeps the coefficient positive and avoids a sign slip.
A function that undoes itself: f(x) = 6x, x ≠ 0
f⁻¹(x) = 6x
the same rule
Check: f(2) = 3 and f(3) = 2 ✓. Its graph is symmetric about y = x, so reflecting it changes nothing.
7
Inverses are how conversions run both ways.
p(d) = 58d
dollars into pesos
d(p) = p58
pesos back into dollars
d(p(30)) = d(1740) = 30 ✓
round trip returns the start
f⁻¹(x) = x + 63 for the opening problem
On the exam
Periodical exams ask for the inverse of a linear or simple rational function, a one-to-one decision, or the domain of the inverse; entrance tests like the value question — given f(a) = b, find f⁻¹(b) — and the reciprocal trap. Swap x and y, then undo the operations in reverse order. For a 10-second check, pick an input, run it through f, then through your f⁻¹: you must land back where you started.
Remember the flow
Check one-to-one · y = f(x) · Swap x and y · Solve for y · Compose to confirm
f⁻¹ means "undo". It never means "one over".
