**Problem**: (Problem-Solving Through Problems - 6.9.12) Suppose that $ f $ is differentiable, and that $ f^{\prime}(x) $ is strictly increasing for $ x \ge 0 $. If $ f(0) = 0 $, prove that $ \frac{f(x)}{x} $ is strictly increasing for $ x > 0 $.

**Solution**: Let $ g(x) $ be the average value of $ f^{\prime}(x) $ over the interval $ [0, x] $. We have

$ \displaystyle g(x) = \frac{1}{x} \int_0^x f^{\prime}(t) dt = \frac{f(x)}{x} $.

But since $ f^{\prime}(x) $ is strictly increasing, as $ x $ increases we are adding values of $ f $ that are larger than the average, so the average will consequently be strictly increasing as well. Hence $ g(x) = \frac{f(x)}{x} $ is strictly increasing for $ x > 0 $, as desired. QED.

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Comment: Noticing the average value function was a nice way to solve this problem and made it "intuitively" clear. We took advantage of the fact that the integral of the derivative is the function itself (the Fundamental Theorem of Calculus).

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Practice Problem: (Problem-Solving Through Problems - 6.9.8) Let $ f: [0,1] \to (0,1) $ be continuous. Show that the equation

$ \displaystyle 2x-\int_0^x f(t)dt = 1 $

has exactly one solution in the interval $ [0, 1] $.

Let F(x) = \int_{0}^{x} f(t) dt. F(x) is a function [0, 1] --> [0, 1), and moreover F(0) = 0.

ReplyDelete2x - F(x) - 1 takes on the value -1 at 0 and a positive value at 1, so it has at least one root. Its derivative, 2 - f(x), is always positive, so it is monotonically increasing; hence, it has exactly one root.

(It seems a little too easy...)

Yeah, it did seem a little easy, but oh well.

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