Rolle and the mean value theorem

Try

A position function moves from 10 metres to 130 metres over eight seconds. Its average velocity is 15 metres per second. Under suitable smoothness assumptions, must it have that instantaneous velocity somewhere? Does this imply it moves at 15 metres per second throughout the interval?

Predict

Sketch a curve connecting two points and the secant line between them. Predict where a tangent parallel to the secant might occur. Then sketch |x| on [−1,1]. The endpoints have equal heights, but neither straight branch has zero slope. Identify the hypothesis this corner will violate.

Experiment

For f(x)=x² on [1,5], the secant slope is (25−1)/(5−1)=6. Since f′(x)=2x, the matching tangent occurs at c=3. Draw the parabola, its secant and the tangent at (3,9).

Check	Value or conclusion
Continuity on [1,5]	polynomial, so yes
Differentiability on (1,5)	polynomial, so yes
Secant slope	6
Solve f′(c)=6	c=3
Interior location	1<3<5

Now replace x² with |x| on [−1,1]. Continuity and equal endpoint values still hold. Differentiability fails at the interior point zero, explaining why the conclusion is not guaranteed.

Understand

Rolle’s theorem states that if f is continuous on a closed interval [a,b], differentiable on its interior and f(a)=f(b), then some interior c has f′(c)=0. The interval has distinct endpoints a<b. One proof uses the extreme value theorem: a continuous function attains maximum and minimum values. If it is constant, every interior point works. If it is nonconstant with equal endpoint values, at least one extreme occurs inside. At a differentiable interior extremum, left and right difference-quotient signs force the derivative to be zero.

The mean value theorem, or MVT, removes the equal-height requirement. Under the same continuity and differentiability hypotheses, some c in (a,b) satisfies f′(c)=(f(b)−f(a))/(b−a). Subtract the secant line to prove it: with m equal to the secant slope, define g(x)=f(x)−f(a)−m(x−a). Then g(a)=g(b)=0 and g inherits the regularity assumptions. Rolle supplies g′(c)=0, which means f′(c)=m.

The theorem guarantees existence, not uniqueness or a midpoint location. In the quadratic example c happens to be the midpoint; that pattern does not hold for arbitrary functions. Nor does knowing an average velocity establish a constant velocity profile. A differentiable position model meeting the endpoint data must reach 15 metres per second at least once, but may vary widely otherwise.

Derivative bounds yield useful global-within-an-interval conclusions. If |f′|≤3 throughout an interval, MVT gives |f(b)−f(a)|≤3|b−a| for any two points there. Similarly, if f′ is zero throughout an interval, any pair of values agrees, so f is constant. The connected interval matters: a function can have zero derivative separately on two disjoint intervals and different constant values on each.

Apply

Write theorem solutions in this order: specify the function and interval; verify each hypothesis; compute or describe the secant slope; state the interior existence conclusion. For the proof exercise, show why the auxiliary function has equal endpoint values. For a derivative-bound task, report an upper bound, not an asserted exact change.

Continue

Next use derivative signs to justify increase, decrease and extrema. MVT explains why a positive derivative throughout an interval implies increasing values there. Its hypotheses prevent conclusions across missing domain points or corners where the required argument changes.