**Problem**: (Problem-Solving Through Problems - 6.5.6) Suppose that $ f: [0,1] \to \mathbb{R} $ is differentiable, $ f(0) = 0 $, and $ f(x) > 0 $ for $ x \in (0,1) $. Let $ k $ be a positive integer. Prove that there is a number $ c $ in $ (0,1) $ such that

$ \frac{kf^{\prime}(c)}{f(c)} = \frac{f^{\prime}(1-c)}{f(1-c)} $. [Generalized]

**Solution**: Consider the function $ g(x) = f^k(x)f(1-x) $; it is clearly also differentiable. Note that $ g(0) = g(1) = 0 $. Thus by Rolle's Theorem we know that there exists a $ c \in (0,1) $ such that $ g^{\prime}(c) = 0 $. But then

$ g^{\prime}(c) = kf^{k-1}(c)f^{\prime}(c)f(1-c)-f^k(c)f^{\prime}(1-c) = 0 $

$ f^{k-1}(c)[kf^{\prime}(c)f(1-c)-f(c)f^{\prime}(1-c) = 0 $.

Since $ f(c) > 0 \Rightarrow f^{k-1}(c) > 0 $, we must have

$ kf^{\prime}(c)f(1-c)-f(c)f^{\prime}(1-c) = 0 \Rightarrow \frac{kf^{\prime}(c)}{f(c)} = \frac{f^{\prime}(1-c)}{f(1-c)} $

as desired. QED.

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Comment: Inspiration for this method comes from cross multiplying and realizing that it looks a lot like the product rule; in fact, for $ k = 1 $ it is just that. Then taking the fact that $ f(x) > 0 $ for $ x \in (0,1) $ allows us to multiply by $ f(c) $ any number of times to obtain the necessary function for integration.

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Practice Problem: (Problem-Solving Through Problems - 6.5.11) If $ f: \mathbb{R} \to \mathbb{R} $ is a differentiable function and $ a $ is an aribtrary real, prove there is a root of $ f^{\prime}(x)-af(x) = 0 $ between any two roots of $ f(x) = 0 $.

Between two consecutive roots x_0, x_1, we must have f'(x_0) and f'(x_1) of opposite signs. (Because the integral of f'(x) across that range is zero.)

ReplyDeleteHence g(x) = f'(x) - a f(x) has opposite signs at the endpoints of the interval, and hence there must be a root.

(I don't know why it took me that long to do this problem =X )