Find f⁻¹(x) algebraically for linear, quadratic, rational, radical, exponential, and logarithmic functions. See every swap-and-solve step, then verify (f∘f⁻¹)(x) = x and watch the reflection symmetry on the graph.
If f(a) = b, then f⁻¹(b) = a — inputs and outputs swap
Reverse mapping: Where f takes x to y, the inverse f⁻¹ takes y back to x. The two functions "undo" each other, so (f∘f⁻¹)(x) = x and (f⁻¹∘f)(x) = x.
Horizontal Line Test: A function has an inverse (is injective) if and only if every horizontal line crosses its graph at most once. Monotone functions always pass this test.
Graphically: The graph of f⁻¹ is the reflection of f over the line y = x. That is why the two curves are mirror images across the diagonal — exactly what the graph above shows.
Notation: f⁻¹(x) does NOT mean 1/f(x). The superscript −1 denotes the inverse function, not a reciprocal.
f(x) = x² has no inverse on ℝ, but does on x ≥ 0
Why quadratics need restriction: f(x) = x² gives f(2) = 4 and f(−2) = 4 — two inputs share one output, so the inverse would need to map 4 to two values, which is not a function. Restricting to x ≥ 0 forces injectivity.
Choosing the branch: By convention, when x ≥ 0 is required, the inverse is the positive square root: f⁻¹(x) = √((x − b)/a). A different restriction (x ≤ 0) would give the negative branch.
Other non-injective types: Sine and cosine need restriction too (e.g. sin on [−π/2, π/2]). Rational functions with equal-degree numerator and denominator can also require care.
Range becomes domain: Always state the domain of f⁻¹ (= range of f). For f(x) = x² + b (x ≥ 0) the range is [b, ∞), so the domain of f⁻¹ is x ≥ b.
One-on-one Precalculus tutoring walks through the swap-and-solve method, domain restrictions, and composition verification on your exact homework — building the fluency that makes inverses feel intuitive.