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Let $f:[a,b] \rightarrow \mathbb{R_{+}^{*}}$ and $f$ derivative in $[a,b]$. Show that: $\exists c\in (a,b)$:

$$\frac{f\left ( b \right )}{f\left ( a \right )}=\frac{{f}'\left ( c \right )}{f\left ( c \right )}e^{b-a}$$

I tried my best and arrived here : $${\ln [f(c)]}'-\frac{f(b)}{e^{b}}\frac{1}{\frac{f(a)}{e^{a}}}=0$$

I would love to hear some ideas or hints. Thanks for your attention!

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    By derivable you mean differentiable? – 2017-01-04
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    I'm just curious why you don't want to use the symbol $[a,b]$ – 2017-01-04
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    @LoganLuther admit a derived function – 2017-01-04
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    @juniven satisfied now ? – 2017-01-04
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    Yes I am, thank you. – 2017-01-04
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    Apply mean value theorem on $\log f(x)$ in $[a, b]$. – 2017-01-04
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    There appears to be a typo in your question. The expression should be $$\frac{f(b)}{f(a)} = \exp\left((b - a)\cdot\frac{f'(c)}{f(c)}\right)$$ The question as it stands is wrong and one can check by putting $f(x) = 1$ when LHS is $1$ and RHS is $0$. – 2017-01-04
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    @ParamanandSingh sorry to disappoint you but what i typed is completly correct – 2017-01-04
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    It may be the given statement, but Paramanand Singh gave you a simple counterexample to show that it cannot be _true_. – 2017-01-04
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    Ok then the typo is not from your side, but it is fault of the book. There can be typos and mistakes in books too and luckily mathematics is a subject where you can ensure (without much effort) that there is a mistake / typo in book. – 2017-01-04

2 Answers 2

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I had an answer using the same hint as suggested by Paramanand Singh in the comments but that doesn't give the required result. He points out that this is probably a typo, so I'll post this suggestion anyway.

You can derive the Mean Value Theorem (MVT) from Rolle's theorem. Apply it to the function $g$ where $g(x) = \ln f(x)$ on $[a,b]$ to get, for a certain $c \in \; ]a,b[$: $$g(b)-g(a) = (b-a)g'(c) \implies \ln f(b) - \ln f(a) = (b-a)\frac{f'(c)}{f(c)}$$ Now the left-hand side is $\ln\tfrac{f(b)}{f(a)}$, so applying the exponential function to both sides gives: $$\frac{f(b)}{f(a)} = \exp\left( (b-a)\frac{f'(c)}{f(c)} \right)$$ Which is not what you wrote, but what might be the intented answer.

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    i wish it was a typo but its not . what i typed there is completly correct – 2017-01-04
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    It may be the given statement, but Paramanand Singh gave you a simple counterexample (the constant function $f(x)=1$) to show that it cannot be _true_. – 2017-01-04
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    oh well i don't know at that point . so what you mean is my question has no solution – 2017-01-04
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    I mean the statement as given is _false_, so it cannot be proven. – 2017-01-04
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You should attempt a rearrangement of the statement that allows you to see where the mean value theorem is clearly used.

At hand, we are given to show that $\frac {f(b)}{f(a)} = \frac{f'(c)}{f(c)} e^{b-a}$. Taking logarithms, we see that $\log f(b)- \log f(a) = (b-a) + \log f'(c)-\log f(c)$. We are tempted to use the mean value theorem for $\log f$ then. But then we don't get the answer you have given!

Let $g(x) = \log f(x)$. Note that by the Lagrange Mean Value theorem, there exists a constant $c$ such that $g'(c) = \frac{g(b)-g(a)}{b-a}$. I leave you to expand and check that the answer given in Paramanand Singh's comment above is the correct one.

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    I checked again the book and what i typed is correct 100% and i really wished it was like what @Paramanad suggested but it's not – 2017-01-04
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    I think there is a typo in your book (as pointed above), in which case, none of us share the blame. Nevertheless, at least you know how to approach this question. – 2017-01-05