The Mean Value Theorem
Theorem 1.1 (Mean Value Theorem) - If f: ℝn → ℝ is differentiable, then for every x,y ∈ ℝn there exists z ∈ [x,y] such that
f(y) − f(x) = ∇f(z) ⋅ (y − x)
Proof - Set φ(t) = f(x + t(y − x)). By proposition 1.2 by applying the chain rule we obtain:
φ′(t) = ∇f(x + t(y − x)) ⋅ (y − x)
It results that φ: [0, 1] → ℝ is differentiable in ]0, 1[ and continuous in [0, 1] with φ(0) = f(x), φ(1) = f(y). By the one-dimensional mean value theorem there exists t ∈ (0,1) such that
φ(1) − φ (0) = φ′(t )(1 − 0)
or equivalently,
f(y) − f(x) = ∇f(z) ⋅ (y − x)
where z = x + t (y − x).□