Inverse Functions

Core Idea

The inverse function f^-1 "undoes" f: if f(a) = b, then f^-1(b) = a. Graphically, f and f^-1 are reflections over the line y = x. A function has an inverse if and only if it is one-to-one (passes the horizontal line test). To find f^-1 algebraically: swap x and y in y = f(x), then solve for y. The composition f(f^-1(x)) = x and f^-1(f(x)) = x verifies the inverse relationship.

How It's Best Learned

Start with simple examples: if f(x) = 2x + 3, find f^-1(x) by swapping and solving. Verify by composition. Use the horizontal line test to determine invertibility. Graph functions and their inverses to see the y = x reflection. Discuss restricting domains to create invertible functions (e.g., restricting x² to x >= 0).

Common Misconceptions

• Confusing f^-1(x) with 1/f(x) (inverse function vs. reciprocal).
• Thinking every function has an inverse (only one-to-one functions do).
• Forgetting to swap x and y before solving.
• Not restricting the domain when needed (e.g., finding the inverse of x² without specifying x >= 0).

Explainer

Think of a function as a machine: you put in an input, it produces an output. The inverse function is the reverse machine — you give it the output and it tells you what the original input was. If f(3) = 7, then f⁻¹(7) = 3. If f(x) = 2x + 3 converts Fahrenheit to some scale, then f⁻¹ converts back. The relationship is captured by the composition rules: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x — doing f then undoing it returns you exactly where you started.

Not every function has an inverse, and this is where the one-to-one (or injective) requirement comes in. A function fails to have an inverse when two different inputs produce the same output — because then the reverse machine wouldn't know which original input to return. For example, f(x) = x² sends both 3 and −3 to 9. If you ask "what input gave output 9?", the function can't answer uniquely. The horizontal line test detects this visually: if any horizontal line crosses the graph more than once, the function is not one-to-one and has no inverse on that domain. You can *create* an inverse by restricting the domain — for x² restricted to x ≥ 0, the inverse is √x.

To find the inverse algebraically, you're essentially solving for the input in terms of the output. Start with y = f(x), swap x and y to get x = f(y), then solve for y. The swap step encodes the reversal: x (the original output) is now the input, and y (the original input) is now the output. For f(x) = 2x + 3: swap to get x = 2y + 3, solve to get y = (x − 3)/2. So f⁻¹(x) = (x − 3)/2. Always verify by composing: f(f⁻¹(x)) = 2·((x − 3)/2) + 3 = (x − 3) + 3 = x. ✓

Graphically, f and f⁻¹ are reflections over the line y = x. This is because swapping x and y in the equation is exactly what reflecting over y = x does to a graph. If you plot f(x) = 2x + 3 and f⁻¹(x) = (x − 3)/2 on the same axes, they are mirror images across the diagonal line. This geometric picture connects the algebra to a visual structure and reinforces why domain restrictions matter: if you restrict x² to x ≥ 0, its graph is the right half of the parabola, and its reflection over y = x is the upper half of √x — they fit together perfectly.