Cauchy's Mean Value Theorem

On the parametric curve (g(t),f(t))(g(t), f(t)) a tangent parallel to the chord exists. For f(t)=t2f(t) = t^2, g(t)=t3g(t) = t^3 it sits at ξ=14/9\xi = 14/9.

f′(ξ)g′(ξ)=f(b)−f(a)g(b)−g(a)\frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}
ξ=149≈1.556\xi = \frac{14}{9} \approx 1.556

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Examples & Exercises

Prove the ordinary Mean Value Theorem is Cauchy's theorem with g(x)=xg(x) = x.

f(b)−f(a)g(b)−g(a)=f′(ξ)g′(ξ)→g(x)=xf′(ξ)=f(b)−f(a)b−a\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)} \xrightarrow{g(x)=x} f'(\xi) = \frac{f(b)-f(a)}{b-a}
ξ=14/9≈1.56\xi = 14/9 \approx 1.56: the tangent ratio 2/(3ξ)=3/72/(3\xi) = 3/7 matches the chord slope.
Show proof
  1. Cauchy's theorem: if f,gf, g are continuous on [a,b][a, b], differentiable on (a,b)(a, b) with g′≠0g' \neq 0, then some ξ∈(a,b)\xi \in (a, b) satisfies f(b)−f(a)g(b)−g(a)=f′(ξ)g′(ξ)\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)}.
  2. Substituting g(x)=xg(x) = x gives g(b)−g(a)=b−ag(b) - g(a) = b - a and g′(ξ)=1g'(\xi) = 1, so the formula becomes f(b)−f(a)b−a=f′(ξ)1=f′(ξ)\frac{f(b)-f(a)}{b-a} = \frac{f'(\xi)}{1} = f'(\xi).
  3. That is exactly the ordinary Mean Value Theorem — the Cauchy form with the special choice g(x)=xg(x) = x.

Prove lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1 with Cauchy's Mean Value Theorem.

sin⁡x−sin⁡0x−0=cos⁡ξx1,0<ξx<x  ⇒  cos⁡ξx→1\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1}, \quad 0 < \xi_x < x \;\Rightarrow\; \cos \xi_x \to 1
ξ=14/9≈1.56\xi = 14/9 \approx 1.56 — the theorem point where the ratio equals the chord slope 37\frac37.
Show proof
  1. For x>0x > 0, sin⁡xx=sin⁡x−sin⁡0x−0\frac{\sin x}{x} = \frac{\sin x - \sin 0}{x - 0}. Applying Cauchy's theorem to f(t)=sin⁡tf(t) = \sin t, g(t)=tg(t) = t on [0,x][0, x] (with g′=1≠0g' = 1 \neq 0), some ξx∈(0,x)\xi_x \in (0, x) gives sin⁡x−sin⁡0x−0=cos⁡ξx1=cos⁡ξx\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1} = \cos \xi_x.
  2. Since 0<ξx<x0 < \xi_x < x, squeezing gives ξx→0\xi_x \to 0 as x→0+x \to 0^+. Cosine is continuous (differentiable implies continuous), so cos⁡ξx→cos⁡0=1\cos \xi_x \to \cos 0 = 1.
  3. For x<0x < 0, the same theorem on [x,0][x, 0] yields ξx∈(x,0)\xi_x \in (x, 0) and likewise cos⁡ξx→1\cos \xi_x \to 1. Hence lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1.

Prove lim⁡x→0ex−1−xx2=12\lim_{x \to 0}\frac{e^x - 1 - x}{x^2} = \frac12 by applying Cauchy's theorem twice. (State the 0/00/0 hypotheses.)

ex−1−xx2=eξ−12ξ=eξ22→x→0+12\frac{e^x-1-x}{x^2} = \frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} \xrightarrow{x \to 0^+} \frac12
ξ=14/9≈1.56\xi = 14/9 \approx 1.56 — note there is only one theorem point, because the ratio 2/(3t)2/(3t) is strictly decreasing.
Show proof
  1. Put f(x)=ex−1−xf(x) = e^x - 1 - x and g(x)=x2g(x) = x^2: then f(0)=g(0)=0f(0) = g(0) = 0, both are continuous on [0,x][0, x] (for x>0x > 0), differentiable on (0,x)(0, x), and g′(t)=2t≠0g'(t) = 2t \neq 0.
  2. Cauchy's theorem gives f(x)g(x)=f′(ξ)g′(ξ)=eξ−12ξ\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)} = \frac{e^{\xi}-1}{2\xi} for some ξ∈(0,x)\xi \in (0, x). Applying it again to f1(t)=et−1f_1(t) = e^t - 1, g1(t)=2tg_1(t) = 2t on [0,ξ][0, \xi] (with f1(0)=g1(0)=0f_1(0) = g_1(0) = 0 and g1′=2≠0g_1' = 2 \neq 0) yields eξ−12ξ=eξ22\frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} for some ξ2∈(0,ξ)\xi_2 \in (0, \xi).
  3. Since 0<ξ2<ξ<x0 < \xi_2 < \xi < x, x→0+x \to 0^+ forces ξ2→0\xi_2 \to 0, and continuity of ete^t gives eξ2→1e^{\xi_2} \to 1. So the ratio tends to 12\frac12. The same argument on [x,0][x, 0] covers x<0x < 0, hence lim⁡x→0ex−1−xx2=12\lim_{x \to 0}\frac{e^x-1-x}{x^2} = \frac12.

9. Differentiation