Mean Value Theorem Preview

From Derivative Bounds to an Exact Slope

The previous tutorial showed how bounds on a derivative control the slopes between pairs of points. For example, if \(m\leq f'(t)\leq M\) throughout an interval, then every secant slope lies between \(m\) and \(M\). A natural next question is whether a secant slope must actually occur as a derivative somewhere between its two endpoints. The Mean Value Theorem answers yes.

Its conclusion is stronger than a bound: for two distinct points, it identifies an interior point at which the instantaneous rate of change exactly equals the average rate of change across the interval. We will first prove Rolle’s Theorem, the special case where the endpoint values agree, and then use it to prove the Mean Value Theorem. The argument relies on the Extreme Value Theorem and the one-sided derivative signs at a local extremum, justified directly below.

Rolle’s Theorem: Equal Endpoint Values

When the values at the ends of an interval agree, the graph either stays at that height or must rise above it or fall below it somewhere in between. In the latter cases, continuity ensures that an interior maximum or minimum is attained. At such an interior extremum, differentiability forces the derivative to be zero.

Proof. By the Extreme Value Theorem, \(f\) attains a maximum and a minimum on the compact interval \([a,b]\). If \(f\) is constant, then \(f'(c)=0\) for every \(c\in(a,b)\), so the conclusion holds.

Now suppose \(f\) is not constant. Since \(f(a)=f(b)\), there must be some \(x_0\in(a,b)\) with \(f(x_0)\ne f(a)\). If \(f(x_0)>f(a)\), the maximum value of \(f\) is greater than the values at both endpoints. It is therefore attained at an interior point \(c\in(a,b)\). At this interior local maximum, for small \(h>0\), \((f(c+h)-f(c))/h\leq0\), while for small \(h<0\) the quotient is \(\geq0\). Since the two one-sided limits agree by differentiability, \(f'(c)=0\).

If instead \(f(x_0)<f(a)\), the minimum value is less than the values at both endpoints and is attained at an interior point \(c\in(a,b)\). The function \(-f\) has a local maximum at \(c\), so the same one-sided derivative result gives \((-f)'(c)=0\), and hence \(f'(c)=0\). In either case, the required point exists. \(\square\)

The hypotheses do distinct jobs: continuity on the closed interval guarantees that an extreme value is attained, while differentiability on the open interval lets us use the derivative at an interior extremum. No derivative at either endpoint is required.

The Mean Value Theorem

To turn Rolle’s Theorem into a statement about a general secant slope, subtract the line joining the graph’s endpoint values. The difference between the function and that line has equal values at the endpoints. Rolle’s Theorem then supplies an interior point where the derivative of this difference is zero.

Proof. Define the affine function \(L:[a,b]\to\mathbb{R}\) by

This line agrees with \(f\) at both endpoints: \(L(a)=f(a)\), and substituting \(x=b\) gives \(L(b)=f(a)+f(b)-f(a)=f(b)\). Set \(g(x)=f(x)-L(x)\). The function \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and its endpoint values satisfy \(g(a)=g(b)=0\). Rolle’s Theorem gives a point \(c\in(a,b)\) such that \(g'(c)=0\).

Since \(L'(x)=(f(b)-f(a))/(b-a)\), the Sum Rule gives

Rearranging yields the stated equality. \(\square\)

Geometrically, the theorem says that somewhere in the interior, the tangent slope equals the slope of the secant line joining the endpoints. The tangent and secant need not be the same line; the conclusion concerns their slopes. The proof makes this precise by subtracting the secant line and applying Rolle’s Theorem to the resulting function.

A Consequence: Zero Derivative Forces Constancy

The Mean Value Theorem also gives a useful conclusion in the opposite direction: if the derivative vanishes at every interior point, no two function values can differ. This consequence turns local derivative information into a global statement about the whole interval.

Proof. Choose any \(x,y\in I\) with \(x<y\). The closed segment \([x,y]\) lies in \(I\). The function is continuous on that segment and differentiable on \((x,y)\), so the Mean Value Theorem gives some \(c\in(x,y)\) such that

By hypothesis \(f'(c)=0\), and therefore \(f(y)-f(x)=0\). Thus \(f(y)=f(x)\) for every ordered pair \(x<y\) in \(I\), which means that \(f\) is constant. \(\square\)

In particular, if \(f\) is differentiable on an open interval and \(f'\) is zero throughout it, then \(f\) is constant there. The interval condition matters: it ensures that the segment between any two points remains in the domain, so the Mean Value Theorem can be applied to that segment.

What the Theorem Does—and Does Not—Say

The Mean Value Theorem requires continuity at both endpoints of the chosen closed interval and differentiability at every interior point. It does not require differentiability at the endpoints. Conversely, having a derivative at most interior points is not enough if differentiability fails at even one point of the open interval: the theorem’s hypotheses must hold throughout the interval.

The conclusion is an existence statement. It guarantees at least one interior point with the specified derivative, but it may be difficult to identify that point explicitly, and there may be several. The quartic example found one point by solving an equation; the theorem itself guarantees existence without requiring a formula for that point.

Finally, the Mean Value Theorem is consistent with the derivative bounds from the previous tutorial. If the derivative stays between \(m\) and \(M\), the theorem identifies a derivative equal to the secant slope, so that slope must also lie between \(m\) and \(M\). The earlier derivative-bounds argument established the bound by comparing monotone auxiliary functions. The Mean Value Theorem offers a different viewpoint: it represents each secant slope as an actual derivative value. This preview establishes the theorem and its proof; later results will use it to draw further conclusions about derivatives and extrema.