Condition 1 fails (not continuous on [a,b]): f(x) = 1/x on [-1, 1], f(-1) = -1, f(1) = 1. Wait — f(a) != f(b) here. Better: f(x) = [x] on [0, 1]. f(0) = 0, f(1) = 1 but f is not continuous at x = 1 (from the left). Even if f(a)=f(b), a discontinuity can prevent the extremum from yielding f'(c) = 0.

Condition 2 fails (not differentiable on (a,b)): f(x) = |x| on [-1, 1]. f(-1) = f(1) = 1, f is continuous. But f'(0) DNE. The minimum is at x = 0 (a corner), not a point of zero derivative.

Condition 3 fails (f(a) != f(b)): f(x) = x on [0, 1]. f'(x) = 1 != 0 for all x. No horizontal tangent exists.

Key exam trap: Students often forget to verify differentiability. The question may give a piecewise function that is continuous but has a corner in (a,b).