# When is $f(x) = \frac{ax+a}{e^{-b}-{(e^{c})}^x}$ monotonic? [duplicate]

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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, ThomasJan 16 '13 at 14:43

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

$$\frac{ax+a}{e^{-b}-{(e^{c})}^x}=\frac{ax}{e^{-b}-{(e^{c})}^x}+\frac{a}{e^{-b}-e^{cx}}$$
and the rightmost expression's derivative is always positive (with $\,a\,>0\,$ , of course)...!