The Rolle's Chain: If f(x) = 0 has n real roots, then:

• f'(x) = 0 has at least n-1 real roots (Rolle's between consecutive roots)
• f''(x) = 0 has at least n-2 real roots
• f^(k)(x) = 0 has at least n-k real roots

Contrapositive (bounding roots): If f'(x) = 0 has at most m roots, then f(x) = 0 has at most m+1 roots.

Example: f(x) = $x^5$ - 20x + 5. f'(x) = 5$x^4$ - 20 = 5($x^4$ - 4) = 5($x^{2-2}$)($x^{2+2}$). f'(x) = 0 at x = +/-sqrt(2) (2 roots). So f has at most 3 real roots.

Proving exactly one root: Show f' > 0 (or f' < 0) everywhere, so f is strictly monotone => at most one root. Then use IVT (sign change) for existence.