Exponential decay function multiplied by a lineair function I have the following two function, one is exponential and the other linear. When i multiply them shouldn't there be an optimum? (So basically a minimum) 

 A: Well, if we multiply an exponential of the form $ae^{bx}$, $a, b \ne 0$, and a linear function of the form $cx + d$, $c \ne 0$, together we get
$f(x) = ae^{bx}(cx + d), \tag{1}$
whence
$f'(x) = abe^{bx}(cx + d) + ace^{bx}, \tag{2}$
and if we set this equal to $0$, we see that $ae^{bx} \ne 0$ cancels out, yielding
$b(cx + d) + c = 0, \tag{3}$
a linear equation with solution
$x_0 = -\dfrac{1}{b} - \dfrac{d}{c} = -\dfrac{bd+ c}{bc}; \tag{4}$
note that by assumption $bc \ne 0$.  $x_0$ is the unique ctitical point of $f(x)$; what kind of critical point is it?  To investigate, we take the second derivative of $f$:
$f''(x) = ab^2e^{bx}(cx + d) + abce^{bx} + abce^{bx}$
$ = ab^2e^{bx}(cx + d) + 2abce^{bx}.  \tag{5}$
We note that $e^{bx} > 0$, always; thus the sign of $f''(x)$ is the same as that of $ab^2(cx + d) + 2abc$.  At $x = x_0$, we have
$cx_0 + d = -\dfrac{c}{b}, \tag{6}$
which follows directly from (3).  Combining these observations, we conclude that the sign of $f''(x_0)$ is the same as that of
$ab^2(cx_0 + d) + 2abc = -abc + 2abc = abc. \tag{7}$
It is in fact easy to write down $f''(x_0)$ in a similar manner, via (5):
$f''(x_0) = abce^{bx_0}, \tag{8}$
From our hypothesis, $abc \ne 0$; the "optimum" $x_0$ is a minimum when $abc >0$ and a maximum when $abc < 0$; there are never points of inflection as long as $abc \ne 0$.  We may infer from the graphs given by EssexPN in the question that $a, c > 0$ and $b < 0$ in the case at hand; $x_0$ is a local maximum; this is in accord with the plot given by Jack in his answer.
A: $$ \frac{d}{dx} a xe^{-bx} = ae^{-cx}(1-bx)$$
This equals $0$ when $x = \frac{1}{b}$
Here's what the plot would look like 

