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Prove: $x \leq f(x) \leq 2x, \forall x\geq0$

conditions:

  1. $f$ is differentiable

  2. $f(0) = 0$

  3. 1 \leq f'(x) \le 2, \forall x\ge0

I've tried to do it by limit defn but couldn't seem to get to the right solution:

$ 1 \le \lim_{x \to c} \frac{f(x)-f(c)}{x-c} \le 2$

how do i manipulate them in such a way that I get $x \leq f(x) \leq 2x, \forall x\geq0 $

I've also noticed that $f(x)$ is an increasing function as f'(x) > 0. Is this information of any use?

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    You can use the code `\to` to obtain $\to$ and replace `->`2012-04-11

3 Answers 3

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I would apply monotony of the defined integral in the close interval [0,x] to the three functions in the inequality number 3 : the constant function 1 , f' , and the constant function 2.

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    Yes, you are right...only we know that the derivate is bounded. and if it had discontinuties they would be from second type because it is the derivate function..so I don't think it can be solved in this way.Thank you.2012-04-11
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Hint: You can study $g(x)=f(x)-x$ and $h(x)=2x-f(x)$, computing their derivatives.

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The key here is to use the Mean Value Theorem. Suppose $f(x)=kx>2x$ for some $x$. Then by the MVT we have some $c\in [0,x]$ such that f'(c)=\frac{f(x)-f(0)}{x-0}=k>2, contradicting the fact that f'(c)\leq 2 for all $c\geq 0$. Similarly, if $f(x)=kx for some $x$ then by the MVT we have some $c\in [0,x]$ such that f'(c)=\frac{f(x)-f(0)}{x-0}=k<1, contradicting the fact that f'(c)\geq 1 for all $c\geq 0$.

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    i managed to obtain a direct proof using t.b.'s hint, with help from some of your workings, thanks!2012-04-11