Deep Dive
Setting up Cauchy's theorem as an extension of Lagrange
Dr. Anuj Kumar begins by anchoring Cauchy's Mean Value Theorem (CMVT) to the prior theorems: Rolle's Theorem applies when f(a) = f(b), and Lagrange's MVT handles cases where f(a) ≠ f(b). Cauchy's theorem takes the next step—instead of one function, you apply Lagrange's formula to two separate functions f(x) and g(x), derive the Lagrange condition for each, then divide the two results together. This eliminates the (b-a) term from both numerator and denominator, yielding the core CMVT formula: f'(c)/g'(c) = [f(b) - f(a)]/[g(b) - g(a)]. The three conditions are: f and g continuous on [a,b], both differentiable on (a,b), and critically, g'(x) ≠ 0 throughout (a,b). If g'(c) were zero, the left side becomes undefined and the theorem collapses.
Verifying CMVT with polynomials x² and x³
The first worked example has f(x) = x² and g(x) = x³ on [1, 2]. Both are polynomials, so continuity on the closed interval is immediate—no verification needed beyond stating the fact. Differentiating yields f'(x) = 2x and g'(x) = 3x². On the open interval (1,2), both derivatives exist and are finite; g'(x) = 3x² is never zero because x = 0 falls outside [1,2]. All three conditions satisfied, Kumar applies the formula: 2c/3c² = (4-1)/(8-1) = 3/7. Simplifying gives c/(3c²) becomes 1/3c = 3/7, so c = 14/9 ≈ 1.56, which indeed lies in (1,2). The verification concludes by confirming c belongs to the open interval and CMVT holds.
Exponential functions and the arithmetic mean result
The second example uses f(x) = e^x and g(x) = e^(-x) on the general interval [a,b]. Both exponential functions are continuous everywhere and differentiable everywhere; their derivatives e^x and -e^(-x) are always finite and never zero on any real interval. Setting up the CMVT equation: e^c / (-e^(-c)) = (e^b - e^a) / (e^(-b) - e^(-a)). Through algebraic manipulation—factoring out negatives, finding common denominators, and canceling terms—the powers combine to yield e^(2c) = e^(a+b). Equating exponents gives 2c = a + b, so c = (a+b)/2. This is precisely the arithmetic mean of the interval endpoints, a result that holds for any choice of a and b. Kumar emphasizes this demonstrates how the theorem's structure forces c to be the midpoint—an elegant relationship between the theorem's conclusion and the interval itself.