The theorem
If is continuous on , differentiable on and , then there is at least one with .
What you are looking at
The curve is on . The first factor makes the ends easy: for every , so the chord is horizontal. The theorem promises a in between with ; the working lists all of them, and the green lines mark the horizontal tangents.
Every condition matters, and the second function shows the most common failure. has equal heights at and is continuous, but it has a corner at : it is not differentiable there, and no point has a horizontal tangent. Rolle’s theorem is the special case of the mean value theorem in which the chord is horizontal.
Try this
- 1Slide from to : the single moves to the right, and once passes a second appears on the left.
- 2Drag along the curve: the tangent is horizontal exactly at the in the working.
- 3Drag or so that : the theorem no longer applies, although a horizontal tangent may still be there.
- 4Switch to : equal heights and continuous, but the corner at breaks differentiability — there is no .
Mistakes students make
- Checking but not differentiability on : a single corner or cusp is enough to lose .
- Asking for too much: differentiability is only needed on the open interval. on satisfies the theorem although it is not differentiable at .
- Expecting exactly one : the theorem gives at least one.
In the exam
- The classic use: prove an equation has at most one root (or at most ). If it had two, Rolle’s theorem would give a zero of that cannot exist.
- To show there is a with some property, build the function: find with whose derivative is the expression you want, and apply the theorem to .
Where it is taught
- AP Calculus AB/BC · Unit 5 (5.1, the Mean Value Theorem, of which Rolle’s is the special case)
- Greek Γ΄ Λυκείου · §2.5
Teaching this?
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