Cauchy (Mean Value Theorem)
1. Statement
If two functions f and g satisfy
- f and g are continuous on the closed interval [a, b],
- f and g are differentiable on the open interval (a, b), and
- g′(x) ≠ 0 for every x in (a, b) — so that the denominators below are non-zero,
then there exists at least one point c in (a, b) such that
Known as Cauchy's Mean Value Theorem (or the generalised mean value theorem). It is the most general of the three theorems: setting g(x) = x gives g′(x) = 1 and the ratio collapses to f′(c) = [f(b) − f(a)]/(b − a) — Lagrange's theorem; and forcing f(a) = f(b) as well gives Rolle's theorem.
Why g(b) ≠ g(a): if g(b) were equal to g(a), Rolle's theorem applied to g would give a point with g′ = 0 inside (a, b), contradicting hypothesis 3. So the denominator cannot vanish.
2. Proof (the auxiliary-function trick again)
Choose the constant λ so that the function F(x) = f(x) − λ g(x) has equal endpoint values:
F is continuous on [a, b], differentiable on (a, b), and F(a) = F(b), so by Rolle's theorem there is c in (a, b) with F′(c) = 0, i.e.
3. Geometric Meaning
Think of the pair (g(t), f(t)) as a parametric curve traced as t runs from a to b. The left side is the slope of the chord joining the two endpoint points of the curve; the right side is the slope of the tangent at t = c. So: on a parametric curve, some tangent is parallel to the chord — exactly the same geometric idea as Lagrange's theorem, but stated for curves that are not graphs of functions.
Verify Cauchy's MVT for f(x) = x² , g(x) = x³ on [1, 2]
Left side = [f(2) - f(1)] / [g(2) - g(1)] = (4 - 1)/(8 - 1) = 3/7
Right side = f′(c)/g′(c) = 2c / 3c² = 2/(3c)
Equate: 2/(3c) = 3/7 ⇒ 14 = 9c ⇒ c = 14/9 = 1.5556
c = 1.5556 lies in (1, 2) ✓ ⇒ verified.
7. Why Cauchy's Form Is Needed — L'Hôpital's Rule
Cauchy's MVT is the exact tool behind L'Hôpital's rule. For a 0/0 form, write F(x) = f(x)/g(x) = [f(x) − f(a)]/[g(x) − g(a)] (using f(a) = g(a) = 0). Cauchy's theorem applied on [a, x] gives
so as x → a (and hence c → a), the limit of the quotient equals the limit of the quotient of derivatives — provided that limit exists. This is the standard proof, and it is the answer to "why can we differentiate numerator and denominator?".
8. The Three Mean Value Theorems Together
| Theorem | Conditions | Conclusion | Special case of |
|---|---|---|---|
| Rolle's | f continuous on [a,b], differentiable on (a,b), f(a) = f(b) | f′(c) = 0 for some c in (a,b) | Lagrange's with f(a) = f(b) |
| Lagrange's | f continuous on [a,b], differentiable on (a,b) | f′(c) = [f(b) − f(a)]/(b − a) | Cauchy's with g(x) = x |
| Cauchy's | f, g continuous on [a,b], differentiable on (a,b), g′ ≠ 0 in (a,b) | [f(b) − f(a)]/[g(b) − g(a)] = f′(c)/g′(c) | the general form |
Uses: Rolle → root separation, constant-function test; Lagrange → monotonicity, inequalities, error bounds, Taylor's theorem, linear approximation; Cauchy → L'Hôpital's rule and Taylor's theorem with Cauchy's remainder.
9. Exam Points
- State Cauchy's Mean Value Theorem and give its conditions, including g′(x) ≠ 0 — 3 marks, almost certain.
- Prove Cauchy's MVT using Rolle's theorem — 5 marks, very frequent.
- Verify Cauchy's theorem for a given pair f, g and find c — 5-7 marks, the most frequent numerical question of this topic.
- Show that Lagrange's theorem is a special case of Cauchy's (take g(x) = x), and that Rolle's is a special case of Lagrange's — 5 marks.
- Explain how Cauchy's MVT is used to prove L'Hôpital's rule — 5 marks, frequent.
- Compare the three mean value theorems in a table — 3-5 marks.