I am trying to show that
|f(x)−f(y)|<|x−y|,
for the function f to be defined as f:[0,+∞)↦[0,+∞), f(x)=(1+x2)1/2, using the mean value theorem.
I have done this:
Since f is differentiable on [0,+∞), then there is a point x0, $x
f(x)−f(y)=(x−y)f′(x0),
by the mean value theorem. Hence,
|f(x)−f(y)|=|x−y||f′(x0)|=|x−y||x0(1+x02)−1/2|≤|x−y||x0|≤|x−y|M<|x−y|
where M is a constant.
Can someone tell me if this is correct?
Answer
Hint:
First prove the general result: if f:R→R is a differentiable function and |f′(x)|<M for all x∈R then for all x,y∈R the inequality |f(x)−f(y)|<M|x−y|
holds. The proof is very similar to what you have done in the question.Next prove that if f(x)=√1+x2 then |f′(x)|<1. To do this consider f′(x)2=x21+x2.
Combinding the two results above gives the desired result.
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