Rolle’s theorem and mean value theorem
“As for my methods, I have sought to give them all the rigour which is demanded in geometry…”
What an average rate forces
A trip covering \(120\text{ km}\) in two hours has average speed \(60\text{ km/h}\). Under a continuous differentiable position model for travel in one direction, there must be an instant when the velocity equals that average. The velocity may reach that value more than once; the total distance and time do not tell us when.
The mean value theorem makes this connection for any differentiable function. We first prove the equal-endpoint case.
Theorem
Rolle's theorem
Suppose \(a<b\), \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(f(a)=f(b)\). Then at least one \(c\in(a,b)\) satisfies \(f^{\prime}(c)=0\).
Proof (optional)
The extreme-value theorem gives an attained maximum and minimum. If \(f\) is constant, its derivative is zero throughout \((a,b)\). Otherwise at least one extremum value differs from the common endpoint value, so it must occur at an interior input. Fermat’s theorem makes the derivative there zero.
A graph that returns to its starting height must either be flat throughout or have an interior highest or lowest point. Differentiability turns that interior turn into a horizontal tangent.
Theorem
Mean value theorem
If \(a<b\), \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then some \(c\in(a,b)\) satisfies
\[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}.\]Geometrically, an interior tangent is parallel to the secant through the endpoints.
Proof (optional)
Let \(m=(f(b)-f(a))/(b-a)\) and subtract the secant line:
\[\begin{aligned} g(x)&=f(x)-f(a)\\ &\quad-m(x-a). \end{aligned}\]This function is continuous on \([a,b]\) and differentiable inside, with \(g(a)=g(b)=0\). Rolle’s theorem gives \(g^{\prime}(c)=0\). Since \(g^{\prime}(x)=f^{\prime}(x)-m\), the desired equality follows.
The theorem is an existence statement. Solving for the point may be easy, difficult, or impossible in elementary formulas. It does not assert a unique point, nor does it assert that the matching tangent is halfway across the interval for a general function.
Example 1
All matching tangents
For \(f(x)=x^3\) on \([-1,1]\), the hypotheses hold and the average slope is
\[\frac{1-(-1)}{1-(-1)}=1.\]Solve \(f^{\prime}(c)=3c^2=1\) to get \(c=\pm1/\sqrt3\). Both lie inside the interval. The theorem promised at least one point; here there are two.
A bound for actual change
If \(m\le f^{\prime}(x)\le M\) throughout \((a,b)\) and the theorem’s hypotheses hold, its conclusion gives
\[\begin{aligned} m(b-a)&\le f(b)-f(a),\\ f(b)-f(a)&\le M(b-a). \end{aligned}\]In particular, \(\lvert f^{\prime}(x)\rvert\le K\) implies \(\lvert f(b)-f(a)\rvert\le K(b-a)\). A bound on the rate becomes a bound on the accumulated change.
Example 2
Certify a change without calculating the endpoint
For \(f(x)=\sqrt{x}\) on \([4,4.1]\),
\[0<f^{\prime}(x)=\frac1{2\sqrt{x}}\le\frac14.\]The theorem gives \(0<\sqrt{4.1}-2\le0.025\). Thus \(2<\sqrt{4.1}\le2.025\). The tangent estimate from the first lesson is now a certified upper bound for this particular value, because we bounded the derivative over the whole interval.
Consequence
A zero derivative on an interval
Let \(f\) be continuous on an interval \(I\) and differentiable in its interior. If \(f^{\prime}(x)=0\) throughout the interior, then \(f\) is constant on \(I\). Continuity at any included endpoints is part of the hypothesis.
Proof (optional)
For any two inputs \(u<v\) in \(I\), the hypotheses hold on \([u,v]\). The mean value theorem yields \(f(v)-f(u)=f^{\prime}(c)(v-u)=0\). Hence every pair of values agrees.
Example 3
Why a corner invalidates the argument
For \(f(x)=\lvert x\rvert\) on \([-1,1]\) the endpoint values agree, so the secant slope is zero. Yet every existing interior derivative is either \(-1\) or \(1\). At zero the derivative is missing, so neither Rolle’s theorem nor the mean value theorem applies.
Checking hypotheses identifies the precise obstruction. Merely having a continuous graph is not enough.
Warning
A zero derivative at one point says nothing about constancy nearby. A zero derivative at every point of an interval does. On a disconnected domain, the constants on different pieces may differ.