Roll’s Theorem

We note here that if \(f(x)=ax+b\), then \(f(x)-f(x_0)=a(x-x_0)\) and so \([f(x)-f(x_0)]/[x-x_0]=a\), and so \(f'(x)=a\) for every \(x\).

Let \(f\) be a derivable function on a segment \(A=[a,b]\), and assume that \(f(a)=f(b)\), then there is a number \(c\) such that \(a<c<b\) and \(f'(c)=0\). Why?

Since \(f\) is derivable, it is continuous. So it attains its minimum \(m\) and its maximum, \(M\). If \(m=M\) then all values of \(f\) are the same, and a constant function has derivative \(0\) at all points. Otherwise, on e of those is not equal to \(f(a)\). The point where the one not equal to \(f(a)\) is attained, is not \(b\), and so it is a \(c\) such that \(a<c<b\). But at this point, a minimum or a maximum is attained not on an endpoint, so that \(f'(c)=0\).

Now let \(f\) be any derivable function on \([a,b]\). Define \(S=[f(b)-f(a)]/[b-a]\), the slope of \(f\). Define \(g\) by \(g(x)=f(x)-[Sx-Sa]\). Then \(g(a)=f(a)\), and \(g(b)=f(b)-Sb=f(b)-S[b-a]=f(b)-[f(b)-f(a)]/[b-a][b-a]=f(b)-[f(b)-f(a)]=f(b)-f(b)+f(a)=f(a)\). Therefore, by the previous discussion, there is \(c\), \(g'(c)=0\). But \(g'(x)=f'(x)-S\), and so \(f'(c)=S=[f(b)-f(a)]/[b-a]\). So we see that if a function is derivable on an interval there is a point where the derivative is equal to the slope.

Here is a simple use: if \(f'(x)\geq 0\) for every \(x\) if \(a<x<b\), then \(f\) is increasing. In order to see why, let aleq cf(c). For similar reasons, if the derivative is always negative, the function is decreasing.

Another way to write the same equation, by the way, is \(f(b)=f(a)+f'(c)[b-a]\) for some \(c\), \(a<c<b\).