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I have the function $f(x) = a x e^{1+ax}$ and I want to find where it has a min or max value.

To do this I calculate the derivative $f'(x) = a^{2}x e^{1+ax}$. This is equal to $0$ only if $a=0$ or $x=0$.

How to proceed from here?

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    You are right since e function have only one max/minmium value.2011-12-08
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    Wrong derivative. Use the Product rule.2011-12-08
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    @AndréNicolas Holy! I better go sleep now. Now I realized that I am really tired.2011-12-08

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Since $f$ is only a function of $x$, we will take $a$ to be constant (unless it's a function of $x$, in which case you need to specify that).

Using the product rule, $f'(x)=ae^{1+ax}+a^2xe^{1+ax}=ae^{1+ax}(1+ax)$. From here, I think you can find the critical point of $f$.

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    So $1+ax=0$ means that $a = 0$ or $x = \frac{-1}{a}$. For a>0 we have a minimum and for a<0 we have a maximum.2011-12-08
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    Well, if $a=0$, then $f(x)=0$ for all $x$, and hence every point is a critical point. Since $y=0$ is a rather boring function, we may as well assume that $a\not=0$. Remember, we're treating $f$ as a function of $x$ and we're fixing $a$ (and assuming that $a\not=0$). So, what is the critical point?2011-12-08
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    The critical point is $x = -1/a$.2011-12-08
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    Okay, so is it a minimum, a maximum, or neither?2011-12-08
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    Generally the answer is that there is no min or max. If we specify a>0 we can say that we have a minimum and if a<0, we have a maximum.2011-12-08
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    Yes, that's correct. Technically, whether or not the critical point is a maximum or a minimum depends on the sign of $a$. However, $f$ has exactly one global extremum.2011-12-08
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    Hint:to determine if the critical point is a global minimizer, maximizer, or neither, determine where $f$ is increasing and where $f$ is decreasing.2012-06-08