maeveoneill
Junior Member
- Joined
- Sep 24, 2005
- Messages
- 93
I am trying to graph the equation 1/x(x-3)^2
Y' =x(x-3)^2(0) - (1)2x(x-3)/ [x(x-3)^2]^2
= -2x^2 + 6x/ [x(x-3)^2]^2
0 = -2x^2 + 6x
= x( -2x +6)
x=0, x=3
by testing the intervals i found that the function decrease between -infinity and 0, and between 3 and positive infinity.. and found that function increases between 0 and 3.
now to find points of inflection i have to find y'' and equate it to 0.
could you show me how to simplify the equation for y'' and find x when y'' = 0
Y' =x(x-3)^2(0) - (1)2x(x-3)/ [x(x-3)^2]^2
= -2x^2 + 6x/ [x(x-3)^2]^2
0 = -2x^2 + 6x
= x( -2x +6)
x=0, x=3
by testing the intervals i found that the function decrease between -infinity and 0, and between 3 and positive infinity.. and found that function increases between 0 and 3.
now to find points of inflection i have to find y'' and equate it to 0.
could you show me how to simplify the equation for y'' and find x when y'' = 0
