How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x)=(x^3)/(x^2-4)#?

1 Answer
Apr 14, 2017

The intervals of increasing are #x in (-oo,-sqrt12) uu(sqrt12, +oo)#
The intervals of decreasing are #x in (-sqrt12,-2)uu(-2,+2)uu(2,sqrt12)#

Explanation:

We calculate the first derivative and construct a sign chart.

We need

#(u/v)'=(u'v-uv')/(v^2)#

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

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

#v=x^2-4#, #=>#, #v'=2x#

#f'(x)=(3x^2(x^2-4)-2x*x^3)/(x^2-4)^2#

#=(3x^4-12x^2-2x^4)/(x^2-4)^2#

#=(x^4-12x^2)/(x^2-4)^2#

#=(x^2(x^2-12))/(x^2-4)^2#

#f'(x)=(x^2(x+sqrt12)(x-sqrt12))/((x+2)^2(x-2)^2)#

We construct the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##-sqrt12##color(white)(aaa)##-2##color(white)(aaa)##0##color(white)(aaa)##2##color(white)(aaa)##sqrt12##color(white)(aaa)##+oo#

#color(white)(aaaa)##x+sqrt12##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(a)##+##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+2##color(white)(aaaaaaaaaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(a)##+##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaaaaaaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(a)##+##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-sqrt12##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(a)##-##color(white)(aa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f'(x)##color(white)(aaaaaaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(a)##-##color(white)(aa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaa)##↘##color(white)(a)##↘##color(white)(aa)##↘##color(white)(aaaa)##↗#

The relative maximum is when #x=-sqrt12#

The relative minimum is when #x= sqrt12#

graph{x^3/(x^2-4) [-14.24, 14.24, -7.12, 7.12]}