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Rolle’s theorem

If a smooth curve starts and ends at the same height, somewhere in between its tangent is horizontal. Change the shape with kk, find every ξ\xi, and watch the theorem fail at a corner.

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The theorem

If ff is continuous on [α,β][\alpha, \beta], differentiable on (α,β)(\alpha, \beta) and f(α)=f(β)f(\alpha) = f(\beta), then there is at least one ξ∈(α,β)\xi \in (\alpha, \beta) with f′(ξ)=0f'(\xi) = 0.

What you are looking at

The curve is f(x)=12(4−x2)(1+kx)f(x) = \dfrac{1}{2}(4 - x^2)(1 + kx) on [−2,2][-2, 2]. The first factor makes the ends easy: f(−2)=f(2)=0f(-2) = f(2) = 0 for every kk, so the chord ABAB is horizontal. The theorem promises a ξ\xi in between with f′(ξ)=0f'(\xi) = 0; 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. f(x)=2−∣x∣f(x) = 2 - |x| has equal heights at ±2\pm 2 and is continuous, but it has a corner at 00: 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

  1. 1Slide kk from 00 to 11: the single ξ=0\xi = 0 moves to the right, and once kk passes 0.50.5 a second ξ\xi appears on the left.
  2. 2Drag PP along the curve: the tangent is horizontal exactly at the ξ\xi in the working.
  3. 3Drag AA or BB so that f(α)≠f(β)f(\alpha) \ne f(\beta): the theorem no longer applies, although a horizontal tangent may still be there.
  4. 4Switch to 2−∣x∣2 - |x|: equal heights and continuous, but the corner at 00 breaks differentiability — there is no ξ\xi.

Mistakes students make

  • Checking f(α)=f(β)f(\alpha) = f(\beta) but not differentiability on (α,β)(\alpha, \beta): a single corner or cusp is enough to lose ξ\xi.
  • Asking for too much: differentiability is only needed on the open interval. 1−x2\sqrt{1 - x^2} on [−1,1][-1, 1] satisfies the theorem although it is not differentiable at ±1\pm 1.
  • Expecting exactly one ξ\xi: the theorem gives at least one.

In the exam

  • The classic use: prove an equation has at most one root (or at most nn). If it had two, Rolle’s theorem would give a zero of f′f' that cannot exist.
  • To show there is a ξ\xi with some property, build the function: find gg with g(α)=g(β)g(\alpha) = g(\beta) whose derivative is the expression you want, and apply the theorem to gg.

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

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