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)xcolor(white)(aaaaaaa)-oocolor(white)(aaaa)-sqrt12color(white)(aaa)-2color(white)(aaa)0color(white)(aaa)2color(white)(aaa)sqrt12color(white)(aaa)+oo

color(white)(aaaa)x+sqrt12color(white)(aaaaaaaa)-color(white)(aaaa)+color(white)(aaa)+color(white)(a)+color(white)(aa)+color(white)(aaaa)+

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

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

color(white)(aaaa)x-sqrt12color(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]}