The function that undoes the function — swap, then solve.
Find f⁻¹(x)
f(x) = x + 23
Swap x and y, then solve: x = y + 23 gives 3x = y + 2, so y = 3x − 2. B is the reciprocal, not the inverse — the little −1 is a label, not a power.
Whatever f does, the inverse undoes — in reverse order.
1. add 2
2. divide by 3
1. multiply by 3
2. subtract 2
Notice the order flips as well as the operations.
start
x
f →
then
f(x)
f⁻¹ →
end
x again
Put a number in, run f, then run the inverse — you land back on the number you started with.
y is easier to move around than the label f(x).
f(x) = x + 23
just a name for y
y = x + 23
Nothing has changed yet — only the name.
Trade places.
y = x + 23
x = y + 23
An inverse turns every input into an output and every output back into an input. Swapping the letters says exactly that.
x = y + 23
the swapped equation
3x = y + 2
the fraction is gone
3x − 2 = y
− 2 both sides
Now put the answer the usual way round.
Multiply by 3 first, exactly as Step 1 predicted.
This is also the answer when a question asks for the g with f(g(x)) = x for every x: that g is f⁻¹, found the same way.
f⁻¹(x) = 3x − 2
The little −1 is a label, not a power.
(x + 2) ÷ 3
3x − 2
The two graphs are reflections of each other in the dashed line y = x.
Where does a rising line meet its inverse? On the mirror line y = x, so solve f(x) = x. For f(x) = 3x − 10: 3x − 10 = x gives x = 5, and the graphs cross at (5, 5).
For one value of the inverse we never need its formula. (a) f(x) = 5x − 8: find f⁻¹(12). (b) f(x) = kx − 1 and f⁻¹(5) = 3: find k.
(a) 5x − 8 = 12
which input gives 9?
x = 4
so f⁻¹(12) = 4
(b) f⁻¹(5) = 3 → f(3) = 5
turn it around
3k − 1 = 5 → k = 2
an ordinary equation
f(12) = 52 answers a different question. For the inverse, the given number is the output; we are hunting for the input.
A negative coefficient. Find the inverse of f(x) = 7 − 3x.
x = 7 − 3y
write y, swap
3y = 7 − x
add 3y, take x — keeps y positive
f⁻¹(x) = 7 − x3
check: f(1) = 4, and 7 − 43 = 1 ✓
What is different: dividing by −3 is where the signs go wrong, so move the y term across first.
With slope −1 the same steps undo themselves. For f(x) = 9 − x: x = 9 − y gives y = 9 − x, so f⁻¹(x) = 9 − x. Such a function is its own inverse.
It sends 4 to −1, not back to 1. Every answer of this kind gets the one-point test: if f(1) = 4, then f⁻¹(4) must be 1.
A cube. Find the inverse of f(x) = 2x³ + 5.
x = 2y³ + 5
write y, swap
y³ = x − 52
undo +5, then ×2
f⁻¹(x) = 3√x − 52
the cube is undone last; check: f(1) = 7, 3√22 = 1 ✓
What is different: f cubes FIRST, so the inverse takes the cube root LAST, over the whole fraction.
A square, kept to one half. Find the inverse of f(x) = (x − 3)² for x ≥ 3.
x = (y − 3)²
write y, swap
y − 3 = √x
the + root only: y ≥ 3 in the inverse
f⁻¹(x) = 3 + √x
check: f(5) = 4, and 3 + √4 = 5 ✓
What is different: on all real x there is no inverse — f(1) and f(5) are both 4. The condition x ≥ 3 picks one half, and that picks the sign of the root.
Find the inverse of f(x) = 2x + 1x − 4.
x = 2y + 1y − 4
swap as usual
xy − 4x = 2y + 1
× (y − 4) both sides
xy − 2y = 4x + 1
y terms left, the rest right
y(x − 2) = 4x + 1
factor out y
f⁻¹(x) = 4x + 1x − 2
divide
Check: f(0) = 1−4 = −¼, and 4(−¼) + 1−¼ − 2 = 0−2¼ = 0 ✓. Pattern for ax + bcx + d: the inverse is −dx + bcx − a — the a and d trade places and both change sign. When d = −a the function is its own inverse.
x − 42x + 1 just turns f upside down. f⁻¹ undoes f; 1f does not.
Expect a fractional f to invert, f⁻¹ of one number, or an unknown constant from f⁻¹(b) = a. For a single value, solve f(x) = b instead of finding the formula. For a formula, test one point: if f(p) = q, the answer must send q back to p.
f(f⁻¹(x)) = x
the two undo each other for every x
Find f⁻¹(x) if
f(x) = 4 − x2
Swap and solve: x = 4 − y2 gives y2 = 4 − x, so y = 8 − 2x. A has both signs flipped. Check it with a number: f(2) = 3, and f⁻¹(3) = 8 − 6 = 2. ✓
Hint: the x term is being subtracted here, so watch the signs when you move it across. The 4 comes off first.
