# Let $f$ be differentiable with $f(0)=0$

If I take $f(x)=x$ then the given conditions are satisfied and hence i can cancel out option $D$.Also by$f(x)=x$,we see that option $B$ is satisfied in case we choose $c=1$. By placing $f(x)=x$ in option $A$,we see that $|x| \leq c$for $|x|\geq 1$ which does not seem to be correct.option $C$ is true only when $c=0$.So option $(C)$can not be right for arbitrary choice of $c$.Am I going in the right direction? Thanks in advance for your time.

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$f(x)=x$ satisfies option C with $c=0$. There are easy correct ways to see that C is incorrect. However, $f(x)=x$ is an easy way to see that A is incorrect. I suggest thinking about proving B independently of this process of elimination method of answering exercises which quickly leaves it as the only possibility. Hint: Mean Value Theorem. –  Jonas Meyer Jan 17 '13 at 4:49
Regarding your edit, it doesn't make sense to say that your particular example of $f(x)=x$ proves that the general statement of B holds. To show that B holds you would have to show that all $f$ satisfying the hypotheses satisfy the conclusion of B. In particular, the $c$ might depend on $f$ a priori. Your edited statements about C also don't get past the problem mentioned in my last comment. You have to give a different counterexample, because $f(x)=x$ does satisfy the hypothesis. It is "there exists $c$" not "for all $c$". For example, try nonlinear, or linear with smaller slope. –  Jonas Meyer Jan 17 '13 at 5:06

(A) Counterexample: Let $f(x)=x$, then $f'(x)=1$ for all $x$. But $\sqrt{x}\to\infty$ as $x\to\infty$, there exists $x$ s.t. $$|x|>c\sqrt{|x|}$$ for all $c>0$.
(B) It is true. By mean value theorem, there exists $c\in(0,x)$ s.t. $$f(x)=f'(c)x.$$ and $|f(x)|=|f'(c)||x| \le 1\cdot |x|$ for all $x$. Since $|x|>1$, we get $|x|<|x|^2$. So $|f(x)|<|x|^2$.