Type 1: Direct verification "Verify LMVT for f(x) = $x^2$ on [1,3] and find c." f(3)-f(1) = 8, (3-1) = 2. f'(c) = 2c = 4, c = 2. Check: 2 is in (1,3). Done.

Type 2: Inequality using LMVT "Prove sqrt(26) < 5.1" f(x) = sqrt(x), LMVT on [25,26]: sqrt(26) - 5 = $\frac{1}{2*sqrt(c}$) for c in (25,26). Since c > 25: $\frac{1}{2*sqrt(c}$) < 1/10. So sqrt(26) < 5.1.

Type 3: Root counting "Find the number of real roots of $x^3$ - 3x + 1 = 0." f'(x) = 3$x^2$ - 3 = 3(x-1)(x+1). f(-1) = 3, f(1) = -1. Sign changes: f(-2) = -1 < 0, f(-1) = 3 > 0, f(1) = -1 < 0, f(2) = 3 > 0. Three sign changes => at least 3 roots. Degree 3 => at most 3 roots. Exactly 3 real roots.

Type 4: Conceptual MCQ "If f is differentiable on R and f(0) = 0, f(1) = 1, f(2) = 1, then..." Apply Rolle's on [1,2]: f'(c1) = 0 for some c1 in (1,2). LMVT on [0,1]: f'(c2) = 1 for some c2 in (0,1). By Darboux (IVP of f'), f' takes every value between 0 and 1.