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Prove that there is no function $f$ on $\left[0,2\right]$ such that $f$ is differentiable and has continuous derivative on interval, $f(0)=-1$ and $f(2)= 4$, and $f'(x) \le 2$ for $x \in \left[0,2\right]$.

My professor suggested using the Mean Value Theorem but I am not sure where to begin or how to proceed.

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  • $\begingroup$ If you apply the MVT for $f$ on the interval $[0,2]$ what do you get... $\endgroup$ – passenger Dec 4 '13 at 15:57
  • $\begingroup$ in this question is $f'(x)\leq 2$ for any $x\epsilon[0,2]$ or all $x\epsilon[0,2]$?? $\endgroup$ – jjoyk Dec 5 '13 at 10:17
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The Mean Value Theorem states that if $g$ is a real-valued function that is continuous on $[a,b]$ and differentiable on $(a,b)$ for some $a<b.$ then there is some point $c\in(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

What does the MVT tell us about your given function?

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  • $\begingroup$ Ok so f'(c) = 5/2 > 2 so f does not exist. $\endgroup$ – Pat Green Dec 4 '13 at 16:12
  • $\begingroup$ That's exactly right. $\endgroup$ – Cameron Buie Dec 4 '13 at 16:29

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