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Let $f_k(x)$ be a function defined on $\mathbb{R}$ by $$f_k(x)=\frac{e^{kx}-1}{2e^x}$$ Where $k$ is a real , How can I study according to the values of $k$ the changes of the changes of $f_k$

I thought about using the derivative, and I did and found

$f^{'}_k(x)=\frac{ke^{kx}-e^{kx}+1}{2e^x}$

We need sign of the numerator , I don't know if this is the best way but I used the derivative of the numerator again Which is $(ke^{kx}-e^{kx}+1)^{'}=e^{kx}(k^2-k)$ Which has the sign of $k^2-k$

I did this and got stuck and can't think of another way . Can someone help please

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    $\begingroup$ When you say changes of $f_k$, what do you mean exactly? Do you mean when $f_k$ changes sign? $\endgroup$
    – Keen-ameteur
    Commented Nov 2, 2023 at 12:58
  • $\begingroup$ @Keen-ameteur I mean to find when $f_k$ is decreasing and when it's decreasing $\endgroup$
    – Mostafa dd
    Commented Nov 2, 2023 at 13:32

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You can replace $t=e^{kx}$ and ask when $kt-t+1=0$. You then see that the numerator vanishes only when $e^{kx}= -\frac{1}{k-1}$. This does not happen as $e^{kx}$ only attains positive values. This should show that the numerator stays positive for all $x\in \mathbb{R}$, and the function is increasing.

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  • $\begingroup$ @keenKeen-ameteur but k is a real so if k<1 $\endgroup$
    – Mostafa dd
    Commented Nov 2, 2023 at 13:55
  • $\begingroup$ @Mostafadd It doesn't really matter, as $e^\alpha>0$ for any $\alpha\in \mathbb{R}$. Also you have to consider the case $k=1$ separately as well. $\endgroup$
    – Keen-ameteur
    Commented Nov 2, 2023 at 14:01

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