Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 28

y=root(3)(x^2+1)
a) Asymptotes
Since the function has no undefined point, so it has no vertical asymptote.
For horizontal asymptotes check if at x->+-oo , the function behaves as a line y=mx+b,
Compute lim_(x->+-oo)f(x)/x to find m
lim_(x->+-oo)root(3)(x^2+1)/x=0
Compute lim_(x->+-oo)f(x)-mx to find b,
lim_(x->+-oo)root(3)(x^2+1)-0x=oo
Since the result is not a finite constant , so there is no horizontal asymptote at +-oo
b) Maxima/Minima
y'=(1/3)(x^2+1)^(-2/3)(2x)
y'=(2x)/(3(x^2+1)^(2/3))
Let's find critical numbers by solving x, for y'=0
(2x)/(3(x^2+1)^(2/3))=0 ,rArrx=0
Let's check the sign of y' by plugging test point in the intervals (-oo ,0) and (0,oo )
y'(-1)=-2/(3((-1)^2+1)^(2/3))=-2/(3(2)^(2/3))=-0.42
y'(1)=2/(3(2)^(2/3))=0.42
There is no maximum point.
Minimum point is at x=0
c) Inflection Points
y''=(2/3)((x^2+1)^(2/3)-x(2/3)(x^2+1)^(-1/3)(2x))/(x^2+1)^(4/3)
y''=(2/3)(3(x^2+1)-4x^2)/(3(x^2+1)^(5/3))
y''=(2/9)(3-x^2)/(x^2+1)^(5/3)
Let's find inflection points by solving x, for y''=0
3-x^2=0 , x=+-sqrt(3)
Inflection points are (sqrt(3) ,root(3)(4) ) and (-sqrt(3) ,root(3)(4) )
Graph: Function is plotted in red color and second derivative is plotted in green color.