How do you find the intervals of increasing and decreasing given #y=(3x^2-3)/x^3#?

The domain of #y# is #D_y=RR-{0} #

We calculate the derivative to find the intervals of increasing and decreasing.

Here, we have a fraction of 2 functions

#(u/v)'=(u'v-uv')/(v^2)# and #(x^n)'=nx^(n-1)#

#u=3x^2-3#, #=># #u'=6x#

#v=x^3#, #=>#, #v'=3x^2#

So,

#dy/dx=(6x^4-9x^4+9x^2)/(x^6)#

#=(9x^2-3x^4)/(x^6)=(3x^2(3-x^2))/(x^6)#

#=(3(3-x^2))/x^4#

So,

#dy/dx=0#, when #3-x^2=0#

#x=-sqrt3# and #x=sqrt3#

Now, we can make our sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-sqrt3##color(white)(aaaa)##0##color(white)(aaaa)##sqrt3##color(white)(aaaa)##+oo#

#color(white)(aa)##dy/dx##color(white)(aaaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aa)##∣∣##color(white)(aa)##+##color(white)(aaa)##-#

#color(white)(aa)##y##color(white)(aaaaaaaaaa)##darr##color(white)(aaaaa)##uarr##color(white)(aaa)##∣∣##color(white)(aa)##uarr##color(white)(aaa)##darr#

So,
The intervals of decreasing are #x in ] -oo,-sqrt3 ] uu [sqrt3, oo[ #

The intervals of increasing are #x in [ -sqrt3,0 [ uu ] 0, sqrt3 ] #

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