## Theorem 6.5.9: Mean Value Theorem

*f*is continuous on

*[a, b]*and differentiable on

*(a, b)*, then there exists a number

*c*in

*(a, b)*such that

Iff'(c) =

*f*and

*g*are continuous on

*[a, b]*and differentiable on

*(a, b)*and

*g'(x) # 0*in

*(a, b)*then there exists a number

*c*in

*(a, b)*such that

### Proof:

The first version of the Mean Value theorem is actually Rolle's theorem in disguise. A simple linear function can convert one situation into the other:

**geometric interpretation
of MVT**

We need a linear function (linear so that we can easily compute its derivative) that maps the line through the two points ( a, f(a) ) and ( b, f(b) ) to the points ( a, 0 ) and ( b, 0 ). If we subtract that map from the function we will be in a situation where we can apply Rolle's theorem.

To find the equation of such a line is easy:

- y - f(a) = ( x - a )

Thus, we define the following function:

- h(x) = f(x) - ( x - a ) - f(a)

Then h is differentiable in ( a, b ) with h(a) = h(b) = 0. Therefore, Rolle's theorem guaranties a number c between a and b such that h'(c) = 0. But then

- 0 = h'(c) = f'(c) -

which is exactly what we had to show for the first part.

The second part is very similar. You can fill in the details yourself by considering the function

- h(x) = f(x) - ( g(x) - g(a) ) - f(a)