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Algebra · Functions

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Inverse functions

Why f⁻¹ runs the machine backwards (and is NOT 1/f), the write-swap- rearrange recipe for finding it, and the ff⁻¹(x) = x check that catches every error.

Algebra · Functions

Inverse functions

Why f⁻¹ runs the machine backwards (and is NOT 1/f), the write-swap- rearrange recipe for finding it, and the ff⁻¹(x) = x check that catches every error.

Why it works

The inverse function f−1f^{-1} runs the machine backwards: if ff turns 5 into 11, then f−1f^{-1} turns 11 back into 5. The notation is a whole symbol — f−1(x)f^{-1}(x) is NOT 1f(x)\frac{1}{f(x)} and the −1-1 is not a power; it's the "undo" badge.

The recipe: write, swap, rearrange. For f(x)=3x−4f(x) = 3x - 4:
  1. Write the machine as an equation: y=3x−4y = 3x - 4.
  2. Undo it — make xx the subject: x=y+43x = \frac{y + 4}{3}.
  3. The inverse machine, written in the usual letter:
f−1(x)=x+43f^{-1}(x) = \frac{x + 4}{3}.

Every rearranging skill transfers — including factorising out when the variable appears twice: for f(x)=3x−1x+2f(x) = \frac{3x - 1}{x + 2}, clearing the fraction and collecting gives f−1(x)=1+2x3−xf^{-1}(x) = \frac{1 + 2x}{3 - x}.

The check that never lies: feed one through the other — f−1(f(x))f^{-1}\big(f(x)\big) must return plain xx. Quick numeric version: f(5)=11f(5) = 11, so f−1(11)f^{-1}(11) had better be 5: 11+43=5\frac{11 + 4}{3} = 5. ✓ Ten seconds, and it catches swapped signs, unflipped operations, all of it.

Evaluating an inverse at a number needs no formula-hunting if you think backwards: f−1(20)f^{-1}(20) asks "what input gives 20?" — solve f(x)=20f(x) = 20. With the formula in hand it's substitution as usual.

Some machines are their own undo. f(x)=4−xf(x) = 4 - x: subtracting from 4 twice returns the start — 4−(4−x)=x4 - (4 - x) = x — so f−1(x)=4−x=f(x)f^{-1}(x) = 4 - x = f(x), a self-inverse. (kx\frac{k}{x} behaves the same way.) If your rearranging lands back on the original formula, that's not an error; it's a property worth stating.