I'm going to prove the M.V.T. and use it to prove Rolle's Theorem.
Statement of Mean Value Theorem
- A function f is continuous on a closed interval [a,b] and differentiable on every point in the interval (a,b) such that for a < c (some point on the closed interval) < b and f(b)-f(a)/b-a = f'(c).
I will show that there exists a point c within the interval where f'(c)=0.
Let g(x)=(f(b)-f(a))x-(b-a)f(x) where x is substituted for c.
Note that g'(x)=(f(b)-f(a))-(b-a)f'(x)
Setting the derivative equal to zero leaves f'(x)=f(b)-f(a)/b-a.
Let a=x and g=f such that:
g(a)=(f(b)-f(a))a-(b-a)f(a).
g(a)= a*f(b)-a*f(a)-b*f(a)+a*f(a)
g(a)=a*f(b)-b*f(a)
And
g(b)=(f(b)-f(a))b-(b-a)f(b)
g(b)=b*f(b)-b*f(a)-b*f(b)+a*f(b) g(b)=a*f(b)-b*f(a)
We can conclude that g(a)=g(b).
- If g'(x)=0 on some point in the closed interval, then g'(x) is zero for every point on the interval.
- If g(a) > g(b) on some point x in the closed interval, then c is a point that is a local maximum and by definition the derivative at a local maximum/minimum is 0 concluding that g'(x)=0.
- If g(a) < g(b) on some point x in the closed interval where c is a point at which the function g(x) has a local minimum and g'(x)=0. By going through all three cases (constant, maximum, and minimum) there will always be some point c where f'(c)=0.
Recall that g(a)=g(b). Using this to satisfy the third condition of Rolle's theorem that:
f(b)-f(a)/b-a=f'(c) will be f'(c)=0 due to f(b)=f(a).
How does it look?