The Secant Meets Its Parallel Tangent

Somewhere on the curve, a tangent is parallel to the secant of slope f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}. Drag ξ\xi to find it.

f′(ξ)=f(b)−f(a)b−af'(\xi) = \frac{f(b)-f(a)}{b-a}

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Examples & Exercises

Apply the Mean Value Theorem to f(x)=x2f(x) = x^2 on [0,2][0, 2] and find ξ\xi explicitly.

f(2)−f(0)2−0=2,f′(ξ)=2ξ=2  ⇒  ξ=1\frac{f(2)-f(0)}{2-0} = 2, \quad f'(\xi) = 2\xi = 2 \;\Rightarrow\; \xi = 1
f(x)=x2f(x) = x^2, ξ=1\xi = 1 — the tangent slope is 2, matching the secant.
Show proof
  1. The secant slope is f(2)−f(0)2−0=4−02=2\frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2} = 2.
  2. Since f′(x)=2xf'(x) = 2x, f′(ξ)=2f'(\xi) = 2 gives 2ξ=22\xi = 2, i.e. ξ=1\xi = 1. As 1∈(0,2)1 \in (0, 2), this matches the theorem's conclusion.

When f(a)=f(b)f(a) = f(b), the Mean Value Theorem becomes Rolle's theorem. Check it for f(x)=x3−3xf(x) = x^3 - 3x on [−1,2][-1, 2].

f(−1)=f(2)=2  ⇒  ∃ ξ∈(−1,2):f′(ξ)=0f(-1) = f(2) = 2 \;\Rightarrow\; \exists\,\xi \in (-1,2): f'(\xi) = 0
At ξ≈1\xi \approx 1 the tangent is horizontal — parallel to the secant (slope 0).
Show proof
  1. f(−1)=−1+3=2f(-1) = -1 + 3 = 2 and f(2)=8−6=2f(2) = 8 - 6 = 2, so the secant slope is 2−23=0\frac{2-2}{3} = 0.
  2. f′(x)=3x2−3=0f'(x) = 3x^2 - 3 = 0 gives x=±1x = \pm 1, and the open interval (−1,2)(-1, 2) contains ξ=1\xi = 1. That is Rolle's theorem — the Mean Value Theorem with zero slope.

Prove via the Mean Value Theorem that f′≡0f' \equiv 0 forces ff to be constant.

f(b)−f(a)=f′(ξ)(b−a)=0f(b) - f(a) = f'(\xi)(b-a) = 0
A constant function: both secant and tangent have slope 0.
Show proof
  1. Assume ff is continuous on [a,b][a, b], differentiable on (a,b)(a, b), with f′≡0f' \equiv 0.
  2. By the Mean Value Theorem, some ξ∈(a,b)\xi \in (a, b) gives f(b)−f(a)=f′(ξ)(b−a)=0⋅(b−a)=0f(b) - f(a) = f'(\xi)(b-a) = 0\cdot(b-a) = 0.
  3. Hence f(b)=f(a)f(b) = f(a), and since a,ba, b were arbitrary, ff is constant.

9. Differentiation