Finding Inverses Algebraically

Pre-Calculus · Axiom Academy

LESSON Finding Inverse Functions Algebraically One recipe for every invertible function: write y=f(x) , swap x and y , solve for y , rename it f^ -1 (x) . Finding f^ -1 is a fixed procedure. You "undo" every operation f performs on x by solving for x in terms of y — and the cleanest way to set that up is to swap the variables first. Run the recipe on f(x) = 3x - 5 . Each line is one step; watch how the swap in step 2 is the only conceptual move — the rest is ordinary algebra. Find the inverse of f(x) = 3x - 5 Geometrically, swapping x and y reflects every point across the line y=x . A point on f lands on on f^ -1 — the same point with its coordinates traded. The recipe is identical for — only step 3 gets longer. After the swap, "solve for y " means clear the denominator, distribute, gather every y -term on one side, factor, then divide. The same swap-then-solve recipe handles every invertible family — only the "undo" operation in step 3 changes. Whatever f builds up, f^ -1 takes apart in reverse order. The reason every one of these works is the same: f^ -1 is built to reverse f , so running a value through f and then f^ -1 returns the value you started with — . You've turned "undo the function" into a four-line recipe — and seen that the swap is a reflection across y=x that any invertible family obeys. Scroll up to revisit any step.

This is the written version of the interactive lesson above. See the full Pre-Calculus course.