Strategy 1: Rolle's Chain (Upper bound) If f^(k)(x) = 0 has at most m roots, then f^(k-1)(x) = 0 has at most m+1 roots. Work down from the highest derivative.

Example: f(x) = $x^5$ - 20x + 5. f'(x) = 5$x^4$ - 20. f''(x) = 20$x^3$. f'' = 0 has 1 root (x=0). So f' has at most 2 roots. So f has at most 3 roots.

Strategy 2: IVT (Lower bound) Find values where f changes sign. Each sign change gives at least one root.

Example: f(-3) < 0, f(-1) > 0, f(1) < 0, f(4) > 0. Three sign changes => at least 3 roots.

Strategy 3: Combine for exact count At least n roots (IVT) + at most n roots (Rolle's chain) = exactly n roots.

Strategy 4: Uniqueness via monotonicity If f'(x) > 0 (or < 0) everywhere, f has at most 1 root. Combine with IVT (sign change) for exactly 1 root.

Example: $x^3$ + 3x + 1 = 0. f'(x) = 3$x^2$ + 3 > 0. Strictly increasing, at most 1 root. f(0) = 1 > 0, f(-1) = -3 < 0. By IVT, exactly 1 root in (-1,0).

Strategy 5: Descartes' Rule of Signs The number of positive roots is at most the number of sign changes in the coefficient sequence. This gives quick upper bounds.

Strategy 6: Graphical analysis For equations like $e^x$ = kx, sketch both sides and count intersections. Use derivatives to understand curve shapes.