Winter 1995 #3

Problem Statement: Let f:[0,1]\rightarrow\mathbb{R} be continuous, f(0)=0, and f'(x) exists for every x\in (0,1). Let f'(x) be increasing on (0,1). Show g(x)=\dfrac{f(x)}{x} is increasing on (0,1).

Proof: We wish to show that g(x) is increasing on (0,1) which means that g'(x)>0 for x\in(0,1).

Since we know that f is continuous and differentiable we may use the Mean Value Theorem. Let x\in (0,1), then the MVT implies that there is some c\in (0,x) such that f'(c)=\dfrac{f(x)-f(0)}{x-o}=\dfrac{f(x)}{x}=g(x). Consider g'(x)=f''(c). We are given that f' is increasing on (0,1), this implies that f''(c)>0 for all c\in (0,1). Thus, g'(x)=f''(c)>0. Since x was arbitrary in (0,1) and f''(c)>0 for all c\in (0,1) this implies that g'(x)>0 for every x\in (0,1).

Thus, g(x) is increasing on (0,1).


Reflection: When I first approached this problem I think I was over-thinking it. I got the equality I have above, but I was worried about concluding that g'(x)>0 for all x\in (0,1) since I only showed it was equal to some f'(c). The thing that saves us is that our x was arbitrary, so we can define g(x) for every x\in (0,1) and show that no matter which x we’re looking at g'(x)>0. The only reason we’re able to conclude anything in this proof is because we know f'(x) is increasing. Without that we wouldn’t be able to say anything.

This entry was posted in Analysis, Continuity, Differentiable, Math, MVT. Bookmark the permalink.

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