Two-step strategy for "exactly one root":

Step 1 (Existence — IVT): Find values a, b where f(a) and f(b) have opposite signs. By IVT, f has at least one root in (a,b).

Step 2 (Uniqueness — Rolle's contrapositive): Show f'(x) != 0 for all x (or f'(x) > 0 for all x). If f had two roots alpha, beta, then by Rolle's, f'(c) = 0 for some c between them — contradiction. So f has at most one root.

Combined: At least one + at most one = exactly one root.

Example: f(x) = $e^x$ - 2x - 1. f(0) = 0 — wait, that's a root at 0! f(2) = $e^2$ - 5 > 0, f(-1) = 1/e + 1 > 0. f'(x) = $e^x$ - 2 = 0 at x = ln 2. f''(x) = $e^x$ > 0, so f is convex. f(ln 2) = 2 - 2ln2 - 1 = 1 - 2ln2 < 0 (since ln2 > 0.5). So f has roots where it crosses zero: one at x = 0 and one where f changes from negative to positive. Exactly 2 roots.