May 2000#2

Problem Statement: Use the Mean Value Theorem to show that \sqrt{x+1}<1+\dfrac{x}{2} for all x>0.

Proof: Let f(x)=1+\dfrac{x}{2}-\sqrt{x+1}. We wish to show that f(x)>0 for every x>0. First note that f'(x)=\dfrac{1}{2}-\dfrac{1}{2\sqrt{x+1}}>0 for all x>0. Let x>0 and consider the interval [o,x]. Since f(x) is continuous and differentiable on [0,x] we may apply the Mean Value Theorem. According to the MVT there exists a point c\in(0,x) such that f'(c)=\dfrac{f(x)-f(0)}{x-0}. We know that f(0)=0 so f'(c)=\dfrac{f(x)}{x}. Since we know f'(x)>0 the previous statement implies that 0<f'(c)x=f(x). Since x was an arbitrary number greater than 0 we may conclude that f(x)>0 for every x>0.


Reflection: This one actually took me a while. My first thought was to use the function f(x) as defined above, but instead of looking at the interval [0,x] I looked at an arbitrary interval [a,b]. This allowed me to conclude that f(a)-f(b)>0 but that didn’t give me what I wanted. It was after realizing that f(0)=0 and that I could use 0 as an endpoint of the interval the proof kind of fell out.

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