Right Differentiability at an Endpoint

Detailed Solution

Let L = lim(x→0+) f(x) and M = lim(x→0+) f'(x), where L and M are finite real numbers. Step 1: Construction of the continuous extension. Since lim(x→0+) f(x) = L, define F : [0,1) → R by F(x) = L, if x = 0, F(x) = f(x), if 0 < x < 1. By construction, F is continuous at x = 0. Step 2: Definition of the right-hand derivative. The right-hand derivative of F at zero is F'_+(0) = lim(h→0+) [F(h) - F(0)]/h. Using the definition of F, F'_+(0) = lim(h→0+) [f(h) - L]/h. Step 3: Application of Lagrange Mean Value Theorem. For any h in (0,1), F is continuous on [0,h] and differentiable on (0,h). Therefore, by Lagrange Mean Value Theorem, there exists c_h in (0,h) such that [F(h) - F(0)]/h = F'(c_h). Since c_h > 0, we have [f(h) - L]/h = f'(c_h). Step 4: Taking the limit. Since 0 < c_h < h, by the Squeeze Theorem, lim(h→0+) c_h = 0. Using the given condition, lim(x→0+) f'(x) = M, we obtain lim(h→0+) f'(c_h) = M. Consequently, F'_+(0) = lim(h→0+) [f(h) - L]/h = lim(h→0+) f'(c_h) = M. Therefore, F'_+(0) = lim(x→0+) f'(x). Hence, OPTION D IS CORRECT. Analysis of Options: Option A: FALSE. The given conditions provide information only near x = 0. They do not guarantee a finite continuous extension at x = 1. Counterexample: f(x) = 1/(1-x), 0 < x < 1. Here, lim(x→0+) f(x) = 1 and lim(x→0+) f'(x) = 1. Both limits are finite, but f(x) → infinity as x → 1-. Hence, no continuous extension to [0,1] exists. Option B: FALSE. The limiting value of the derivative need not be zero. Counterexample: f(x) = 3x + 2. Then, lim(x→0+) f'(x) = 3 ≠ 0. Option C: FALSE. The hypotheses guarantee a continuous extension at x = 0, but they do not guarantee a continuous extension at x = 1. The function f(x) = 1/(1-x) provides a counterexample. Option D: TRUE. By the Lagrange Mean Value Theorem, [f(h) - L]/h = f'(c_h), where 0 < c_h < h. As h → 0+, c_h → 0+. Therefore, F'_+(0) = lim(x→0+) f'(x). Thus, Option D is the unique correct answer.