Minimize the function Minimize the function 
$$f(x) = (ax+b)^2 \left(\frac{c}{x} + d\right),$$
where $a , b , c \text{ and } d$ are all positive constants and $x$ is the variable.
Thanks and regards
 A: This is a case in which we can set the derivative equal to zero and solve for x.  
As the comment notes, the function appears to dive to -$\infty$ near zero for x < 0 so some restriction is appropriate. $x > 0?$ 
Without much loss of generality you can see this by letting a = b = c = d = 1. The behaviour of the function will not change much and you can also see how the derivative behaves. There is a local minimum near 1/2 and this corresponds to a point where f'(x) crosses the x axis. 
Then it's just an exercise in finding the derivative with the constants in place, setting the derivative equal to 0 and solving for x. 
Note that when you solve you may get more than one solution. The restriction x > 0 should resolve any doubt about which one corresponds to the desired minimum.
A: After some simplification, the derivative is equal to
$$\frac{ax+b}{x^2}\left(2dx^2+(2-a)cx-bc\right).$$
The $\frac{ax+b}{x^2}$ part is safely positive, so we can concentrate on the quadratic part.  
Since the parameters are positive, the discriminant of the quadratic is positive, so the quadratic has two distinct real roots. Since the constant term is negative, one of the real roots is negative, and hence irrelevant, and the other is positive. 
It is easy to see that the quadratic is negative a little past $0$, because of the $-bc$ constant term. So for a while for $x\gt 0$ our function decreases. It reaches a minimum at the positive root of the quadratic, and then increases. So there is indeed a unique positive $x$ at which our function reaches a minimum. We can find the place explicitly by solving a quadratic equation. 
