If f(x)= (x^2)/(x-2)^2f(x)=x2(x2)2, what are the points of inflection, concavity and critical points?

1 Answer
Jun 16, 2017

The critical point is =(0,0)=(0,0). The inflection point is =(-1,1/9)=(1,19). See below for the concavities.

Explanation:

We need

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

Let's calculate the first derivative

f(x)=x^2/(x-2)^2

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

v=(x-2)^2, =>, v'=2(x-2)

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

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

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

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

f'(x)=0, when x=0

We build our first chart

color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaaaaaa)0color(white)(aaaaaaaaaa)2color(white)(aaaaaa)+oo

color(white)(aaaa)-4xcolor(white)(aaaaaa)+color(white)(aaaa)0color(white)(aaaa)-color(white)(aaaa)||color(white)(aaa)-

color(white)(aaaa)(x-2)^3color(white)(aaa)-color(white)(aaaa)0color(white)(aaaa)-color(white)(aaaa)||color(white)(aaa)+

color(white)(aaaa)f'(x)color(white)(aaaaa)-color(white)(aaaa)0color(white)(aaaa)+color(white)(aaaa)||color(white)(aaa)-

color(white)(aaaa)f(x)color(white)(aaaaaa)color(white)(aaa)0color(white)(aaaa)color(white)(aaa)||color(white)(aaa)

The critical point is when x=0

The intervals of increasing are x in (0,2) and decreasing are x in (-oo,0) uu (2,+oo)

Now, we calculate the second derivative

u=-4x, =>, u'=-4

v=(x-2)^3, =>, v'=3(x-2)^2

So,

f''(x)=(-4(x-2)^3+4x*3(x-2)^2)/(x-2)^6

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

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

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

=(8(x+1))/(x-2)^4

Therefore,

f''(x) =0 when x=-1, this is the inflexion point

The chart is as follows

color(white)(aaaa)Icolor(white)(aaaaaa)(-oo,-1)color(white)(aaaa)(-1,2)color(white)(aaaa)(2,+oo)

color(white)(aaaa)f''(x)color(white)(aaaaaaa)-color(white)(aaaaaaaaa)+color(white)(aaaaaaaa)+

color(white)(aaaa)f''(x)color(white)(aaaaaaa)nncolor(white)(aaaaaaaaa)uucolor(white)(aaaaaaaa)uu

The curve is concave is up when x in (-1,2) uu (2,+oo)

The curve is concave down when x in (-oo,-1)

Also,

f''(0)>0, this is a minimum

graph{x^2/(x-2)^2 [-11.36, 11.14, -1.58, 9.67]}