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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. – Victor Dec 8 '11 at 22:30
Wrong derivative. Use the Product rule. – André Nicolas Dec 8 '11 at 22:42
@AndréNicolas Holy! I better go sleep now. Now I realized that I am really tired. – pressy_paris Dec 8 '11 at 22:44
up vote 0 down vote accepted

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. – pressy_paris Dec 8 '11 at 22:58
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? – user5137 Dec 8 '11 at 23:01
The critical point is $x = -1/a$. – pressy_paris Dec 8 '11 at 23:05
Okay, so is it a minimum, a maximum, or neither? – user5137 Dec 8 '11 at 23:05
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. – pressy_paris Dec 8 '11 at 23:09

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