Derivatives, Rolle and the mean value theorem

Try

For f(x)=x² between x=0 and x=4, the secant slope is 4. The derivative f′(x)=2x equals that slope at x=2. The mean value theorem guarantees such a point for any function continuous on the closed interval and differentiable in its interior. It is an existence theorem about a derivative value, not an assertion that the function follows its secant line everywhere.

Predict

Predict why differentiability alone at the two endpoints would not be enough. The theorem needs local derivative information throughout the interior and continuity connecting that behavior to the endpoint values. The function |x| on [−1,1] is continuous with equal endpoint values, yet its derivative is never zero where it exists. The missing derivative at zero prevents Rolle’s conclusion from being applied.

Experiment

Compare three functions on [−1,1].

Function	Continuous throughout?	Differentiable inside?	Rolle applies?
x²	Yes	Yes	Yes
|x|	Yes	No at zero	No
Constant 2	Yes	Yes	Yes, every interior point works

For x², verify the zero derivative at zero directly from the difference quotient. For absolute value, compute the left and right quotients separately. The table tests hypotheses rather than classifying functions by the smoothness of a rendered curve.

Understand

Differentiability at a implies continuity there: for x≠a, write f(x)−f(a) as its difference quotient times x−a. The first factor tends to finite f′(a) and the second to zero. At an interior local maximum, right-hand difference quotients are nonpositive and left-hand ones nonnegative. If the two-sided derivative exists, its value must be zero; a minimum is analogous.

Rolle’s theorem now follows. A continuous function on [a,b] attains extrema. If it is constant, any interior point has zero derivative. If it is not constant and f(a)=f(b), at least one attained extremum lies in the interior, where the derivative vanishes. To obtain the mean value theorem, subtract the secant line. The resulting function has equal endpoint values, so Rolle supplies c with f′(c)=(f(b)−f(a))/(b−a). All endpoint and differentiability hypotheses are inherited by the auxiliary function.

Apply

A derivative sign yields a global comparison. For arbitrary x<y in an interval, apply MVT on [x,y]. If f′≥0 inside, f(y)−f(x)=f′(c)(y−x)≥0, so f is nondecreasing. A function whose derivative is zero throughout an interval is therefore constant. The interval requirement matters: a function can take different constants on two disjoint intervals while its derivative is zero everywhere on its domain.

Repair “strictly increasing means its derivative is positive everywhere.” The function x³ is strictly increasing but has derivative zero at zero. A strictly positive derivative is sufficient for strict increase; it is not necessary. Similarly, an isolated zero derivative does not force an extremum.

Continue

Optional depth derives a Lipschitz bound from |f′|≤M on an interval, together with the required endpoint continuity: MVT gives |f(y)−f(x)|≤M|y−x|. This connects differentiation with uniform continuity and turns a derivative estimate into a global error bound.

Save a theorem application that checks the closed-interval and open-interior conditions separately. Give one example where a hypothesis fails and explain why the conclusion might then fail, rather than claiming every failed hypothesis forces failure. Next, derivatives will be shown to have intermediate values even when they are discontinuous, revealing a subtle distinction beyond MVT.