Question #53c40

1 Answer
Dec 14, 2016

graph{1/(x(x-3)^2) [-10, 10, -5, 5]}

Explanation:

The domain if the function f(x) = frac 1 (x(x-3)^2)f(x)=1x(x3)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.