Mean Value Theorem

maths ✒ mvt

Suppose:

Then $ \exists \phi \in (a,b) $ s.t. $ f(\phi) = \frac { f(b) - f(a) } {b - a } $

Proof

Let $ g : z \mapsto f(z) - f(a) - { z \cdot \frac { f(b) - f(a) } {b - a } } $

Then by AOL, g is continuous over $ [a,b] $ and differentiable over $ (a,b) $

$ g' : z \mapsto f'(z) - \frac { f(b) - f(a) } {b - a}

Notice that $ g(a) = g(b) = 0 $.

Therefore by Rolle's Theorem, $ \exists \phi \in (a,b) $ s.t. $ g'(\phi) = 0 $

Then by our expression for $ g' $, we conclude that $ f'(\phi) = \frac { f(b) - f(a) } {b - a } $