Inverse functions
The idea
An inverse function undoes another: if f sends 4 to 8, then f⁻¹ sends 8 back to 4. Formally f⁻¹(f(x)) = x for every x in the domain, which means inputs and outputs trade places — the domain of f becomes the range of f⁻¹ and vice versa, and the two graphs are mirror images across the line y = x. Inverses surround you already: subtraction undoes addition, a logarithm undoes an exponential, and converting Celsius back to Fahrenheit undoes the conversion that went the other way.
To find a formula, write y = f(x), swap x and y, and solve for y — the algebra of the solve IS the undoing procedure, executed in reverse order. Not every function qualifies: f must be one-to-one, meaning no two inputs share an output, or the reversal is ambiguous; the horizontal line test checks this on a graph. The misconception lodged inside the notation is reading f⁻¹(x) as 1/f(x). The superscript names the inverse function, not a reciprocal — the inverse of f(x) = x + 5 is x − 5, nothing like 1/(x + 5).
Worked example
Find the inverse of f(x) = 2x/(x − 3), state which value is excluded from the domain of f⁻¹, and verify the inverse using x = 4.
- Write y = 2x/(x − 3) and swap the variables: x = 2y/(y − 3). Solving this for y rebuilds the rule in reverse.
- Clear the fraction by multiplying both sides by (y − 3): xy − 3x = 2y. Collect the y terms on one side: xy − 2y = 3x, so y(x − 2) = 3x.
- Divide by x − 2, which demands x ≠ 2: y = 3x/(x − 2). So f⁻¹(x) = 3x/(x − 2), with x = 2 excluded from its domain.
- That exclusion is no accident: f has horizontal asymptote y = 2, so 2 never appears as an output of f — and a value that f never produces cannot be fed backwards through f⁻¹.
- Verify with a round trip: f(4) = 8/(4 − 3) = 8, and f⁻¹(8) = 24/(8 − 2) = 24/6 = 4. The inverse returns 8 to 4, exactly as required.
Answer. f⁻¹(x) = 3x/(x − 2), defined for all x ≠ 2; the round trip f(4) = 8 and f⁻¹(8) = 4 confirms it.
Check your understanding
- Why must a function pass the horizontal line test before an inverse function can exist, and how can restricting the domain rescue a failing function?
- Why are the graphs of a function and its inverse reflections of each other across the line y = x?
- How does the domain exclusion of f⁻¹ encode the range of f, and why must that connection always hold?
- What goes wrong when someone reads f⁻¹(x) as 1/f(x), and how does the notation invite that mix-up?
Build the foundations first
Inverse functions builds on these concepts. If any feel shaky, start there.