The domain if the function #f(x) = frac 1 (x(x-3)^2)# is the union of the intervals #(-oo,3) uu (0,3) uu (3,+oo)#.
First we analyze the behavior of the function on the boundaries of its domain:
#lim_(x->-oo) frac 1 (x(x-3)^2) = 0#
#lim_(x->+oo) frac 1 (x(x-3)^2) = 0#
#lim_(x->0^-) frac 1 (x(x-3)^2) = -oo#
#lim_(x->0^+) frac 1 (x(x-3)^2) = +oo#
#lim_(x->3) frac 1 (x(x-3)^2) = +oo#
So we can see that #f(x)# has an horizontal asymptote for #y=0# and a vertical asymptotes for #x=0# and #x=3#.
Then we calculate the first derivative using the chain rule:
#f'(x) = -frac 1 (x^2(x-3)^4) ((x-3)^2+ 2x(x-3)) = -frac ((x-3)(x-3+2x)) (x^2(x-3)^4) = -frac (3(x-1)) (x^2(x-3)^3) #
We can see that the only critical point is for #x=1#
The second derivative is extremely complex to calculate, so we rather solve the disequation:
#f'(x) > 0#
around #x=1#.
We see that #f(x) < 0# for #x<1# and #f(x) > 0# for #x>1# so the point #x=1# is a minimum.
