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).

\Box

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.

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

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