Local minimal about x=0 What should be the value of a for wich the function f(x)=$ (x^2-2ax+2a)/(x-a) $  have a local mínimum in x=0?
Derivating (i don´t include the calculus) one a twince i arrive to the answer that this funtion doesn´t has a minimun.
If a=0, the the funtion is reduced to f(x)=$x^2/x$ that is f(x)= x for all x not equal to zero, and indefined if x=0, so, i think, is imposible that f has a local minimun here due to, in x=0 the function is not defined.
The other options that I have are: a=1 and a=2, and i think nethier of them are correct.
Is all correct in my treatment of the problema? 
 A: To determine the appropriate $a$ compute $$f'(x;a)=0\Big|_{x=0}$$
and solve for $a$.
This will insure $x=0$ is a critical point of $f$ from which you can perform the appropriate analysis to see whether this indeed corresponds to a minimum.  If not, the problem has no solution.
A: We can get a pretty fair idea of the function's behavior by writing its expression as
$ \ f(x) \ = \ \frac{x^2 \ - \ 2ax \ + \ 2a}{x \ - \ a} \  = \ \frac{x^2-2ax+a^2}{x-a} \ + \frac{2a \ - \ a^2}{x-a} \ = \ \frac{(x \ - \ a)^2}{x-a} \ + \frac{a \ (2 \ - \ a)}{x-a} $ $  = \ (x \ - \ a) \ + \frac{a \ (2 \ - \ a)}{x-a} \ , \ x \neq a \ . $
We see immediately from this that the graph of $ \ f(x) \ $ for  $ \ a \ = \ 0 \ $ is the line  $ \ y \ = \ x \ $ with a "hole" at the origin, so the function is undefined at  $ \ x \ = \ 0 \ $ , as you observed, and possesses no minimum value.  A similar situation holds for $ \ a \ = \ 2 \ $ , for which the graph is the line $ \ y \ = \ x \ - \ 2 \ $ with a "hole" at  $ \ x \ = \ 2 \ $ .
For all other values of  $ \ a \ $ , the function expression is the sum $ \ (x \ - \ a) \ + \frac{k}{x-a} \ $ , $ \ k \ $ a constant, which has a graph resulting from the sum of a line with a slope of $ \ 1 \ $ and a "diagonal" hyperbola.
In the cases with $ \ a \ < \ 0 \ $ and $ \ a \ > \ 2 \ $ , we have $ \ (x \ - \ a) \ - \frac{k}{x-a} \ , \ k \ > 0 \ $ , so the "branch" of the hyperbola to the "left" of the vertical asymptote is "concave upward" (curves upward) and the branch to the "right" is "concave downward". Alternatively, we can say that both terms is the function expression are always increasing, so $ \ f(x) \ $ has only the critical point at $ \ x \ = \ a \ $ where $ \ f'(x) \ $ is undefined, and so has neither a maximum nor a minimum value.
For $ \ 0 \ < \ a \ < \ 2 \ $  , the expression is $ \ (x \ - \ a) \ + \frac{k}{x-a} \ , \ k \ > 0 \ $ , so the  hyperbola is oriented with  the "left" branch  "concave downward", and the "right" one "upward". But this causes the function to "run toward negative infinity" as $ \ x \ \rightarrow \ a^{-} \ $ and "toward positive infinity" as $ \ x \ \rightarrow \ a^{+} \ $ . We now have critical points, but the one to the "left" of the vertical asymptote is a relative maximum .  For $ \ a \ = \ 1 \ $ , there is such a relative maximum at $ \ x \ = \ 0 \ $ . So we never have a relative minimum at  $ \ x \ = \ 0 \ $ .
If we take the trouble to calculate the first derivative, we obtain $ \ f'(0)  \  = \ \frac{2a \ · \ (a-1)}{a^2} \ $ , which confirms that we can only have a critical point at $ \ x \ = \ 0 \ $ for $ \ a \ = \ 1 \ $ (and that we have a meaningless result for $ \ a \ = \ 0 \ $ ) .
The "geometrical" analysis above is somewhat easier to deal with than computing the second derivative.  If we do so, though, we find that  $ \ f''(0)  \  = \ \frac{2 \ · \ (a-2)}{a^2} \ $ ; this is negative for  $ \ a \ = \ 1 \ $ , confirming a relative maximum.  So your surmise that the function you present has no relative minimum at  $ \ x \ = \ 0 \ $ is correct.
