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I have the following task in my homework:

Let $a,b \in \mathbb{R}, a < b $ and $f:[a,b] \to \mathbb{R} $ be a continuous function that is differentiable on $(a,b)$. Show that if $f'(x)=f(x)$ for all $x \in (a,b)$, then there is a $c \in \mathbb{R}$ with $f(x)=ce^x$ for all $x \in [a,b]$. Hint: Consider the function $x\mapsto f(x)e^{-x} $

I know that the two solutions here would be $c=1$ and $c=0$, but apparently, that has to be shown with the mean value theorem, and that hint just confuses me more. Any help would be appreciated!

pavus
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  • Any $c$ would do, not only $c=1$ and $c=0$. The mean-value theorem is used to show that $f(x)e^{-x}$ is constant. – Martin R Jul 08 '18 at 17:43

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If $g (x)=f (x)e^{-x} $, then $$g'(x)=f (x)e^{-x}-f (x)e^{-x}=0. $$ The role of the Mean Value Theorem is to show that a function with zero derivative is constant.

As mentioned by Ryan, the above shows that $f(x)=ce^x$ on $(a,b)$. As $f$ is assumed continuous on $[a,b]$, the equality extends to $[a,b]$.

Martin Argerami
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  • This will show that $f(x)=ce^x$ on $(a,b)$ and then continuity of $f$ on $[a,b]$ forces $f(x)=ce^x$ on $[a,b]$. –  Jul 08 '18 at 18:01
  • Good point. $ $ – Martin Argerami Jul 08 '18 at 18:04
  • @Ryan: That is not necessary, the mean-value theorem only requires the differentiability in the interior of the interval. – However I wonder why Martin reopened the question, it was asked and answered here: https://math.stackexchange.com/q/58097/42969. – Martin R Jul 08 '18 at 18:06
  • @MartinR: The Mean Value Theorem only requires differentiability in the interior of the interval, as you say, so can only be used to show that $f(x)=ce^x$ in the interior of the interval. The question asks to show that $f(x)=ce^x$ on the entire interval $[a,b]$. It's a minor point, admittedly, but when grading this homework, I would look for at least a mention of what to do with the endpoints. –  Jul 08 '18 at 18:31
  • @Ryan: I disagree. $g$ is continuous on $[a, b]$ and differentiable (with derivative zero) on $(a, b)$. $g$ satisfies the conditions of https://en.wikipedia.org/wiki/Mean_value_theorem, and it follows that $g$ is constant on $[a, b]$ . – Martin R Jul 08 '18 at 18:33
  • @MartinR: I was simply pointing out that the claim of the MVT itself gives that $g$ is constant on $(a,b)$ and the next step is that the continuity of $g$ on all of $[a,b]$ allows one to extend this result to all of $[a,b]$. This is pointed out in the section "A simple application" of the Wikipedia link you just copied. Of course, since continuity is required on all of $[a,b]$ for the MVT, perhaps you're point is that that this may be naturally included in the statement of this corollary. –  Jul 08 '18 at 18:48
  • @Ryan: And I still disagree (and that part in the Wikipedia article is not optimal). For any $x\in (a, b]$ the MVT ensures the existence of some $c \in (a, x)$ such that $g(x) - g(a) = g'(c)\cdot(x-a) = 0$. Thus $g$ is constant on $[a, b]$. There is no need to make this a two-step process. – Martin R Jul 08 '18 at 18:51
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    @MartinR: I stand corrected. There is no need to make this a two-step process, as you show. –  Jul 08 '18 at 18:58