The Unapologetic Mathematician

The Differential Mean Value Theorem

Let’s say we’ve got a function $f$ that’s continuous on the closed interval $\left[a,b\right]$ and differentiable on $(a,b)$. We don’t even assume the function is defined outside the interval, so we can’t really set up the limit for differentiability at the endpoints, but they don’t matter much in the end.

Anyhow, if we look at the graph of $f$ we could just draw a straight line from the point $(a,f(a))$ to the point $(b,f(b))$. The graph itself wanders away from this line and back, but the line tells us that on average we’re moving from $f(a)$ to $f(b)$ at a certain rate — the slope of the line. Since this is an average behavior, sometimes we must be going faster and sometimes slower. The differential mean value theorem says that there’s at least one point where we’re going exactly that fast. Geometrically, this means that the tangent line will be parallel to the secant we drew between the endpoints. In formulas we say there is a point $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$.

First let’s nail down a special case, called “Rolle’s theorem”. If $f(a)=0=f(b)$, we’re asserting that there is some point $c\in(a,b)$ with $f'(c)=0$. Since $\left[a,b\right]$ is compact and $f$ is continuous, the extreme value theorem tells us that $f$ must take a maximum and a minimum. If these are both zero, then we’re looking at the constant function $f(x)=0$, and any point in the middle satisfies $f'(c)=0$. On the other hand, if either the maximum or minimum is nonzero, then we have a local extremum at a point $c\in(a,b)$ where $f$ is differentiable (since it’s differentiable all through the open interval). Now Fermat’s theorem tells us that $f'(c)=0$ since $c$ is a local extremum! Thus Rolle’s theorem is proved.

Now for the general case. Start with the function $f$ and build from it the function $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)$. On the graph, this corresponds to applying an “affine transformation” (which sends straight lines in the plane to other straight lines in the plane) to pull both $f(a)$ and $f(b)$ down to zero. In fact, it’s a straightforward calculation to see that $g(a)=0=g(b)$. Thus Rolle’s theorem applies and we find a point $c$ with $g'(c)=0$. But applying our laws of differentiation, we see that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}$. And so $f'(c)=\frac{f(b)-f(a)}{b-a}$, as desired.

January 22, 2008 Posted by | Analysis, Calculus | 22 Comments