Graphing max/min points of y = 1/x(x-3)^2

[maeveoneill]

I am hoping for some quick help on figuring out max/min points for graphing.

The function is y = 1/x(x-3)^2

To find the max and min points, I need to find the derivative. Could somebody solve this for x? I've tried and gotten stuck. Should I rearrange the equation first?

Thanks

 

[stapel]

Why would you want to solve the function you're graphing for "x="...?

Instead, try differentiating the function, to find y'. Then set this equal to zero, and solve.

If you get stuck, please reply showing everything you have tried so far.

Thank you.

Eliz.

 

[maeveoneill]

sorry, that is what i meant.. to find x when y'' = 0.. but i just figured that outtt

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