13. 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 \[ g'(c)=0. \]

Before reading the proof, look at this plot showing \(4\) continuous functions satisfying \(g(a)=g(b)\):

1) one which has the constant value \(g(a)\) red \(g(x) =g(a)\) 2) one which is always bigger than \(g(a)\) blue \(g(x) \geq g(a)\) 3) one which is always smaller than \(g(a)\), cyan \(g(x) \leq g(a)\) 4) and one which is sometimes bigger and sometimes smaller than \(g(a)\). orange \(g(x)\ \) sometimes \(\ \gt g(a)\) and \(g(x)\ \) sometimes \(\ \lt g(a)\) Notice that at every minimum or maximum, the slope is \(0\).

This plot shows four continuous functions on the interval [a,b].
All of these functions are equal at these endpoints. One is the horizontal
line between the endpoints. Another is always above this line. Another
is always below this line. The last is partly above and partly below.

Since \(g(x)\) is continuous, by the Extreme Value Theorem it must have a maximum and a minimum somewhere in \([a,b]\). Since \(g(a)=g(b)\), either

\(g\) is constant and every point in \((a,b)\) is both a maximum and minimum.
\(g(x) \gt g(a)\) for some point \(x\) in \((a,b)\) and so there is a maximum at some point in \((a,b)\).
Or \(g(x) \lt g(a)\) for some point \(x\) in \((a,b)\) and so there is a minimum at some point 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 (secant) line between the endpoints is \[ y=f_\text{sec}(x)=f(a)+\dfrac{f(b)-f(a)}{b-a}(x-a) \]

The plot shows the graph of a parabola opening downward. The piece
between x = a and x = b is highlighted and there is a secant line between
the endpoints.

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

We define a new function \(g(x)\) which is the difference between our original function \(f(x)\) and the straight line between the endpoints: \[ g(x)=f(x)-f(a)-\dfrac{f(b)-f(a)}{b-a}(x-a) \]

The plot shows the same parabola and secant line. Between them
there are a bynch of vertical lines. The parabola is labeled f of x.
The secant line is labeled f sub sec of x. The difference is labeled
g of x.

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.