Fermat: interior extrema are stationary
Fermat's theorem says: if f has a local maximum or minimum at an interior point c, and f is differentiable at c, then f′(c) = 0. The proof reads the sign of the difference quotient from each side. At a local max the numerator f(x) − f(c) is ≤ 0 on both sides; the denominator changes sign, so the two one-sided limits squeeze f′(c) between ≤ 0 and ≥ 0.
Fermat: f has a local max at interior c, f differentiable at c => f'(c) = 0.
Local max: f(x) <= f(c) for all x near c, so f(x) - f(c) <= 0 near c.
Approach from the RIGHT (x > c, so x - c > 0):
(f(x) - f(c)) / (x - c) = (<=0) / (>0) <= 0,
hence the right-hand limit f'(c) <= 0.
Approach from the LEFT (x < c, so x - c < 0):
(f(x) - f(c)) / (x - c) = (<=0) / (<0) >= 0,
hence the left-hand limit f'(c) >= 0.
f is differentiable at c, so both one-sided limits equal f'(c). Thus
f'(c) <= 0 AND f'(c) >= 0 => f'(c) = 0. ∎Rolle, then the mean value theorem
Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then f′(c) = 0 for some interior c. The proof combines two earlier giants — the extreme value theorem hands us a max and a min, and Fermat finishes the interior case.
- By the extreme value theorem, the continuous f attains a max M and a min m on the compact [a, b].
- If both are attained only at the endpoints, then M = m (since f(a) = f(b)), so f is constant and f′ = 0 everywhere — done.
- Otherwise an extremum is attained at an interior point c. Apply Fermat there: f′(c) = 0.
The mean value theorem is Rolle viewed on a tilt. Subtract off the secant line so the endpoints level out, apply Rolle, and tilt back.
MVT: f continuous on [a,b], differentiable on (a,b)
=> exists c in (a,b) with f'(c) = ( f(b) - f(a) ) / ( b - a ).
Define the 'subtract the secant' auxiliary function:
g(x) = f(x) - [ f(a) + (f(b)-f(a))/(b-a) · (x - a) ].
Check g's endpoints:
g(a) = f(a) - f(a) = 0,
g(b) = f(b) - [ f(a) + (f(b)-f(a)) ] = f(b) - f(b) = 0.
So g(a) = g(b) = 0, and g is continuous on [a,b], differentiable on (a,b).
By Rolle, exists c in (a,b) with g'(c) = 0. But
g'(x) = f'(x) - (f(b)-f(a))/(b-a),
so g'(c) = 0 gives f'(c) = ( f(b) - f(a) ) / ( b - a ). ∎