The domain if the function f(x) = frac 1 (x(x-3)^2)f(x)=1x(x−3)2 is the union of the intervals (-oo,3) uu (0,3) uu (3,+oo)(−∞,3)∪(0,3)∪(3,+∞).
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.