Let $f:[a,b]\rightarrow \mathbb{R}$ be a positive and differentiable function. Prove: $$\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$$
I tried to apply the MVT and make some algebraic manipulations, but got stuck.
Any help appreciated.
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$\begingroup$ Do you want to prove that the equality holds for a certain $c$, a certain $a<c<b$, or any $c$? $\endgroup$– toliveiraMar 8, 2017 at 19:42
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2$\begingroup$ Did you forget an exponential somewhere, e.g. in $e^{\frac{f'(c)}{f(c)}}$? $\endgroup$– Clement C.Mar 8, 2017 at 19:42
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$\begingroup$ @ClementC. Sorry for that, editing. $\endgroup$– Itay4Mar 8, 2017 at 19:44
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$\begingroup$ Guys, he means $e^{(b-a)\frac{f'(c)}{f(c)}}$ $\endgroup$– Gabriel RomonMar 8, 2017 at 19:44
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$\begingroup$ @LeGrandDODOM Yet, that is not what (s)he wrote initially. Hence my comment. $\endgroup$– Clement C.Mar 8, 2017 at 19:46
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2 Answers
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Let $g(x)=\ln(f(x))$. According to the mean value theorem, $$ \frac{g(b)-g(a)}{b-a}=g^{\prime}(c)=\frac{f^{\prime}(c)}{f(c)} $$ for some $c\in(a,b)$. This implies that $$ \ln\Big(\frac{f(b)}{f(a)}\Big)=(b-a)\frac{f^{\prime}(c)}{f(c)} $$ hence $$ \frac{f(b)}{f(a)}=\exp\Big((b-a)\frac{f^{\prime}(c)}{f(c)}\Big) $$
answered Mar 8, 2017 at 19:45
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1$\begingroup$ Darn, you beat me by a few seconds. (+1) $\endgroup$ Mar 8, 2017 at 19:47
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High-level idea: The reflex to use the MVT is a good one. Now, the issue is that you do not want to apply it directly to $f$, but to some other function $g$ related to $f$. In particular, the term $e^{b-a}$ strongly hints that you're going to take an exponential after that; and seeing $\frac{f'(c)}{f(c)}$ should prompt you to think of $(\ln f)' = \frac{f'}{f}$. These two together suggest we set $g=\ln f$, and apply the MVT to $g$: good thing, this is legitimate, as $f$ is assumed positive. So let's try that...
Let $g\colon [a,b]\to \mathbb{R}$ be defined by $g\stackrel{\rm def}{=} \ln f$. $g$ is well-defined, and differentiable, as $f$ is positive and differentiable.
Applying the Mean Value Theorem on $g$, we get that there exists $c\in(a,b)$ such that $$ g(b)-g(a) = g'(c) (b-a) $$ and, exponentiating both sides, we obtain $$ e^{g(b)-g(a)} = e^{g'(c)(b-a)}. $$ Recalling that $e^{g(b)-g(a)} = \frac{e^{g(b)}}{e^{g(a)}} = \frac{f(b)}{f(a)}$ and $g'(c) = \frac{f'(c)}{f(c)}$ concludes the proof.
answered Mar 8, 2017 at 19:46
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2$\begingroup$ The right-hand side of your last displayed equation should be $\exp(g^{\prime}(c)(b-a))$. Great answer otherwise though... $\endgroup$ Mar 8, 2017 at 20:18
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$\begingroup$ @carmichael561 Thanks for spotting this! $\endgroup$ Mar 8, 2017 at 20:19