If y= (x^2+1)/(x^2-4) , what are the points of inflection, concavity and critical points?

1 Answer
Sep 20, 2017

The critical point is at (0, -1/4) and it is a local maximum.
The intervals of convexity are (-oo, -2) uu (2,+oo)
The interval of concavity is (-2,2)

Explanation:

Calculating the first and second derivatives and building the variation charts

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

The domain of y is x in {RR -(-2,2)}

u(x)=x^2+1, =>, u'(x)=2x

v(x)=x^2-4, =>, v'(x)=2x

Therefore,

dy/dx=(u'v-uv')/(v^2)=(2x(x^2-4)-2x(x^2+1))/(x^2-4)^2

=(2x^3-8x-2x^3-2x)/(x^2-4)^2

=(-10x)/(x^2-4)^2

The critical points are when dy/dx=0

=>, -10x=0, x=0

Building the variation chart

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

color(white)(aaaa)dy/dxcolor(white)(aaaaaaa)+color(white)(aaaa)||color(white)(aaa)+color(white)(aaa)0color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)-

color(white)(aaaa)ycolor(white)(aaaaaaaaa)color(white)(aaa)||color(white)(aaa)color(white)(a)-1/4color(white)(aaaa)color(white)(aaa)||color(white)(aaaa)

Calculating the second derivative to determine the concavity and the points of inflections

u(x)=-10x, =>, u'(x)=-10

v(x)=(x^2-4)^2, =>, v'(x)=4x(x^2-4)

(d^2y)/(dx^2)=(u'v-uv')/(v^2)=(-10(x^2-4)^2+40x^2(x^2-4))/(x^2-4)^4

=((x^2-4)(-10x^2+40+40x^2))/(x^2-4)^4

=(10(3x^2+4))/(x^2-4)^3

When x=0, =>, (d^2y)/(dx^2)<0, this is a local max

Building the variation chart

color(white)(aaaa)Intervalcolor(white)(aaaa)(-oo,-2)color(white)(aaaa)(-2,2)color(white)(aaaa)(2,+oo)

color(white)(aaaa)Sign (d^2y)/(dx^2)color(white)(aaaaaaaa)+color(white)(aaaaaaaaaa)-color(white)(aaaaaaaa)+

color(white)(aaaa)ycolor(white)(aaaaaaaaaaaaaaa)uucolor(white)(aaaaaaaaaa)nncolor(white)(aaaaaaaa)uu

See the graph below

graph{(x^2+1)/(x^2-4) [-7.9, 7.9, -3.95, 3.95]}