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I had a rehearsal test today and got this question in the test that completely stumped me.

Let $f: \mathbb{R} \to \mathbb{R}$ be a non constant continuous function such that

$$(e^x-1)f(2x)=(e^{2x}-1)f(x)$$

If $f'(0)=1$, then what are $f(x)$ and $f(2x)$?

My try:

I just differentiated both sides with respect to $x$ and made use of $f'(0)=1$, but that got me nowhere.

Please help me someone. Thanks in advance.

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    $\begingroup$ Does $e^{2x}-1 = (e^{x} - 1)(e^{x} + 1)$ help? $\endgroup$ – mattos May 10 '17 at 13:46
  • $\begingroup$ On using the logic of @Mattos and then differentiating the equation and making use of the fact that $f'(0)=1$, I'm getting $f(0)=0$, but that would mean $f(x)=x$ for every $x$ in the domain? $\endgroup$ – Tanuj May 10 '17 at 13:55
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    $\begingroup$ The easiest way is to just notice from how the $ x $ and $ 2 x $ appear on either side of the equation, that $ f(x) = e^x - 1 $ works. $\endgroup$ – user81327 May 10 '17 at 13:55
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I also attended the same test. Here's how I solved it . Its easier than you think it is. First rearrange both sides $$\frac{f(2x)}{(e^{2x}-1)}= \frac{f(x)}{(e^{x}-1)}$$

From this you can conclude that $f(x)= e^{x}-1$,[NOTE: Not necessarily $e^x-1$. Please refer Daniel Fischers comment below.]

Otherwise the above condition would cease to be true. Also, $f'(0) = 1$ holds true for the above function.

As it is an objective question paper this is the quickest (and apparently the only single method). Now you can solve for the options given in the question paper.

Note: This question was an MCQ in which multiple options are correct.

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  • $\begingroup$ How do you know? Did you give this test too? $\endgroup$ – Tanuj May 10 '17 at 14:00
  • $\begingroup$ Yes @user38227 I gave it too. I also assume that you where pretty confused by seeing the options. $\endgroup$ – Ananth Kamath May 10 '17 at 14:03
  • $\begingroup$ How do you conclude $f(x) = e^x-1$ from $\frac{f(2x)}{e^{2x}-1} = \frac{f(x)}{e^x-1}$? $\endgroup$ – Daniel Fischer May 10 '17 at 14:07
  • $\begingroup$ How did you solve the limits then? It actually is quite confusing. $\endgroup$ – Tanuj May 10 '17 at 14:08
  • $\begingroup$ @DanielFischer I noted that from the equality given, $f(x)$ must be of the form $k(e^x - 1)$, so that the equality holds true.Also $k=1$ from the fact that $f'(0)=1$. This may not be the only solution, of-course.But I have mentioned it as the quickest and simplest one I could find. $\endgroup$ – Ananth Kamath May 10 '17 at 14:15
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Note that if $x\ne 0$, we have $$\frac{f(2x)}{e^{2x}-1}=\frac{f(x)}{e^x-1}$$ So put $g(x)=\dfrac{f(x)}{e^x-1}$ for $x\ne 0$. Then we have $$g(2x)=g(x)$$ for all non-zero $x$. $g$ is continuous on $\mathbb R-\{0\}$, so we must have $$g(x) = \begin{cases} c_+, & \text{if $x > 0$} \\ c_-, & \text{if $x < 0$} \end{cases}$$ for some constants $c_+,c_-$.

(Edited to add: It's not that simple $-$ see Daniel Fischer's comment below.)

This gives us $$f(x) = \begin{cases} c_+(e^x-1), & \text{if $x > 0$} \\ c_-(e^x-1), & \text{if $x < 0$} \end{cases}$$

By continuity of $f$, $c_+=c_-=c,$ say. So $f(x)=c(e^x-1)$ for all $x$.

I will leave the rest to you.

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  • $\begingroup$ From $g(2x) = g(x)$ for $x\neq 0$ it does not follow that $g$ is locally constant. Consider $\sin \bigl(\frac{2\pi}{\log 2} \log \lvert x\rvert\bigr)$. You need the other conditions on $f$ to reach the conclusion. $\endgroup$ – Daniel Fischer May 10 '17 at 14:06
  • $\begingroup$ @DanielFischer: Oh yes, you're quite right! And we even have that $f(x)=(e^x-1)\sin \bigl(\frac{2\pi}{\log 2} \log \lvert x\rvert\bigr)$ is continuous. So we also need the implied condition that $f$ is differentiable at $0$. But how do we use that? This question is difficult! $\endgroup$ – TonyK May 10 '17 at 14:13
  • $\begingroup$ Yes indeed. @DanielFischer I also took up the same a assumption which TonyK used $\endgroup$ – Ananth Kamath May 10 '17 at 14:22

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