Find f⁻¹(x) algebraically and verify with the composition f(f⁻¹(x)) = x.
y = f(x) → swap x,y → solve for y = f⁻¹(x)
1. Replace f(x) with y. Rewrite the function as an equation with y on the left: y = ax + b.
2. Swap x and y. Write x = a(y) + b — this is the key step that finds the reflection.
3. Solve for y. Isolate y using algebra. Every operation done to x in f(x) is now "undone" in the opposite order.
4. Write as f⁻¹(x). Replace y with f⁻¹(x) and note any domain restrictions.
f⁻¹ exists ⟺ f is one-to-one (injective)
A function has an inverse if and only if it is one-to-one: every output corresponds to exactly one input.
Horizontal Line Test: If any horizontal line crosses the graph more than once, the function is not one-to-one and has no inverse over that domain.
Restricted Domain: Non-one-to-one functions (like quadratics) can be made invertible by restricting the domain — for f(x) = a(x−h)² + k we use x ≥ h to keep only the increasing half.
One-on-one Algebra tutoring builds real intuition for why swapping x and y works — and when restricted domains are needed. We work through your actual problems.