# 12. Differentials & Linear Approximation

## c. Mean Value Theorem

## 1. Proof

Before we can prove the Mean Value Theorem, we need to prove Rolle's Theorem.

If \(g(x)\) is a function on the interval \([a,b]\) with a continuous derivative, \(g'(x)\), and \(g(a)=g(b)\) then there is a number \(c\) in \((a,b)\) such that \[ g'(c)=0. \]

Look at this plot showing \(4\) continuous functions satisfying \(g(a)=g(b)\):

one which has the constant value \(g(a)\)
**red**
\(g(x) =g(a)\)
one which is always bigger than \(g(a)\)
**blue**
\(g(x) \geq g(a)\)
one which is always smaller than \(g(a)\),
**cyan**
\(g(x) \leq g(a)\)
and one which is sometimes bigger and sometimes smaller than \(g(a)\).
**orange**
\(g(x)\ \) sometimes \(\ \gt g(a)\)
and \(g(x)\ \) sometimes \(\ \lt g(a)\)

Since \(g(x)\) is continuous, by the Extreme Value Theorem it must have a maximum and a minimum somewhere in \([a,b]\). If \(g\) is constant, then every point is both a maximum and minimum. Since \(g(a)=g(b)\), if \(g\) is not constant, then it must go up or down somewhere (as in the plot) and have a maximum or minimum somewhere in \((a,b)\). Let \(c\) be a number in \((a,b)\) where \(g(c)\) is a maximum or minimum. Since \(g(x) \) is differentiable, the slope of the tangent line is \(0\) at \(x=c\), in other words, \[ g'(c)=0 \]

There may be more than one number \(c\) where \(g'(c)=0\).
The theorem only guarantees that there is *at least* one.

If \(f(x)\) is a function on the interval \([a,b]\) with a continuous derivative, \(f'(x)\), then there is a number \(c\) in \((a,b)\) such that \[ f'(c)=\dfrac{f(b)-f(a)}{b-a} \] or equivalently, \[ f(b)=f(a)+f'(c)(b-a) \]

The equation of the line between the endpoints is \[ y=f(a)+\dfrac{f(b)-f(a)}{b-a}(x-a) \]

#### Why is this a line?

It is a line because \(y\) is a linear function of \(x\). In particular, it is the point-slope form of the line, \(y=f(a)+m(x-a)\), where the slope is \(m=\dfrac{f(b)-f(a)}{b-a}\).

#### Why does it go through the endpoints?

When \(x=a\), \[ y=f(a)+\dfrac{f(b)-f(a)}{b-a}(a-a)=f(a) \] When \(x=b\), \[ y=f(a)+\dfrac{f(b)-f(a)}{b-a}(b-a)=f(b) \]

We define a new function \(g(x)\) which is the difference between our original function \(f(x)\) and the straight line between the endpoints: \[ g(x)=f(x)-f(a)-\dfrac{f(b)-f(a)}{b-a}(x-a) \] We evaluate \(g(x)\) at the endpoints and also compute its derivative: \[\begin{aligned} g(a)&=f(a)-f(a)-\dfrac{f(b)-f(a)}{b-a}(a-a)=0 \\ g(b)&=f(b)-f(a)-\dfrac{f(b)-f(a)}{b-a}(b-a)=0 \\ g'(x)&=f'(x)-\dfrac{f(b)-f(a)}{b-a} \end{aligned}\] Since \(g(a)=g(b)\), we can apply Rolle's Theorem to \(g(x)\) to conclude there is a number \(c\) in \((a,b)\) such that \(g'(c)=0\). But this says: \[ f'(c)-\dfrac{f(b)-f(a)}{b-a}=0 \] which is the Mean Value Theorem.