Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 22

y=x/(1-x^2)
a) Asymptotes
Vertical asymptotes are the zeros of the denominator of the function.
1-x^2=0rArr(1+x)(1-x)=0
x=-1 , x=1
Vertical asymptotes are x=1 and x=-1
Degree of numerator=1
Degree of denominator=2
Since degree of denominator degree of numerator,
So, Horizontal asymptote is x-axis i.e. y=0
b) Maxima/Minima
y'=((1-x^2)-x(-2x))/(1-x^2)^2
y'=(1-x^2+2x^2)/(1-x^2)^2
y'=(1+x^2)/(1-x^2)^2
To find critical numbers , solve for x at y'=0
(1+x^2)/(1-x^2)^2=0 rArr1+x^2=0
x^2=-1rArrx=+-i
Since there is no real solution , so there are no critical points.
Domain : -1< x< 1
Check the sign of y' by plugging test points in the intervals (-oo ,-1),(-1,1) and (1,oo )
y'(-2)=(1+(-2)^2)/(1-(-2)^2)^2=5/9
y'(0.5)=(1+(0.5)^2)/(1-0.5^2)^2=2.22
y'(2)=(1+2^2)/(1-2^2)^2=5/9
There is no change in sign of y' , so there are maxima and minima.
c) Inflection points
y''=((1-x^2)^2(2x)-(1+x^2)(2)(1-x^2)(-2x))/(1-x^2)^4
y''=((1-x^2)((1-x^2)2x+4x(1+x^2)))/(1-x^2)^4
y''=(2x-2x^3+4x+4x^3)/(1-x^2)^3
y''=(2x^3+6x)/(1-x^2)^3
y''=(2x(x^2+3))/(1-x^2)^3
For inflection points y''=0
(2x(x^2+3))/(1-x^2)^3=0
2x(x^2+3)=0
x=0 , x^2=-3
ignore the points that are complex,
So , Inflection point is at x=0