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When is $f(x) = \frac{ax}{e^{-b}-{(e^{c})}^x}$ monotonic?

When is $f_2(x) = \dfrac{ax+a}{e^{-b}-{(e^{c})}^x}$, where $(a,b,c)$ are positive real numbers, monotonic?

Taking the derivative of $f_2(x) = \dfrac{ax+a}{e^{-b}-{(e^{c})}^x}$ I obtain:

$$f'_2=d\frac{a}{e^{-b}-e^{cx}}+\dfrac{ce^{cx}(a+ax)}{(e^{-b}-e^{cx})^2} = \dfrac{a e^b(1+e^{(b+cx)}(cx+c-1))}{(e^{(b+cx)}-1)^2}$$

When is $f'_2$ positive? Attempting to use Reduce in Mathematica fails.

Please note that this question is a follow-up to: When is $f(x) = \frac{ax}{e^{-b}-{(e^{c})}^x}$ monotonic? Which, in fairness, I must see to completion given the fact that there are already three answers. However, I am particularly concerned with this somewhat more complex version of the question.

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marked as duplicate by DonAntonio, Stefan Hansen, rschwieb, Did, Thomas Jan 16 '13 at 14:43

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what makes you think this is different from your previous question? –  nbubis Jan 16 '13 at 13:23
    
@nbubis Because here we have: $f'_2=\frac{a}{e^{-b}-e^{cx}}+\frac{ce^{cx}(a+ax)}{(e^{-b}-e^{cx})^2} = \frac{a e^b(1+e^{(b+cx)}(cx+c-1))}{(e^{(b+cx)}-1)^2}$, and in the previous expression we have: $f'_2=\frac{a}{e^{-b}-e^{cx}}+\frac{ce^{cx}(a+ax)}{(e^{-b}-e^{cx})^2} = \frac{a e^b(1+e^{(b+cx)}(cx-1))}{(e^{(b+cx)}-1)^2}$ –  Harrison Jan 16 '13 at 13:25
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1 Answer 1

What "more complex"? It's almost the same since

$$ \frac{ax+a}{e^{-b}-{(e^{c})}^x}=\frac{ax}{e^{-b}-{(e^{c})}^x}+\frac{a}{e^{-b}-e^{cx}}$$

and the rightmost expression's derivative is always positive (with $\,a\,>0\,$ , of course)...!

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Perhaps I shouldn't have said "more complex", but the derivatives of the two expressions are distinct. –  Harrison Jan 16 '13 at 13:28
    
I think the point is that Harrison isn't quite clear on the first answers, and hence, with this problem. Is that fair, Harrison? If you don't quite follow an answer to an earlier question, you can ask for clarification, and you are perfectly entitled to wait to accept an answer until you are satisfied you understand. –  amWhy Jan 16 '13 at 13:29
    
@amWhy That would probably be a fair characterization. –  Harrison Jan 16 '13 at 13:30
    
I really don't want to make people upset here with similar questions. I'm simply stuck. –  Harrison Jan 16 '13 at 13:30
    
@amWhy Thanks - the answer that I'm really looking for is for this version of the question. And it's not entirely clear to me that the answers are going to be very similar. This was definitely my fault - I posted the question and then was unresponsive for a short while. –  Harrison Jan 16 '13 at 13:33
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