Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 27

y=xsqrt(2+x)
a) Asymptotes
Domain of function is x >= -2
Since the function has no undefined points , so it has no vertical asymptotes.
Horizontal Asymptotes:
Let's find the limits of the function at +-oo
Since -oo is not in the domain so there is no horizontal asymptote at -oo
Compute lim_(x->oo)f(x)/x to find m
lim_(x->oo)(xsqrt(2+x))/x
=lim_(x->oo)sqrt(2+x)=oo
Since the slope is not a finite constant, so there is no horizontal asymptote at +oo
b) Maxima/Minima
y'=x(1/2)(2+x)^(-1/2)+sqrt(2+x)
y'=x/(2sqrt(2+x))+sqrt(2+x)
y'=(x+2(2+x))/(2sqrt(2+x))
y'=(3x+4)/(2sqrt(2+x))
Let's find critical points by solving x for y'=0,
(3x+4)/(2sqrt(2+x))=0
3x+4=0 ,=>x=-4/3
Let's check the sign of y' by plugging test points in the intervals (-2,-4/3) and (-4/3 ,oo )
y'(-1.5)=(3(-1.5)+4)/(2sqrt(2+(-1.5)))=-0.354
y'(0)=(3(0)+4)/(2sqrt(2+0))=4/(2sqrt(2))=sqrt(2)
There is no maxima.
Minimum point is at x=-4/3
y(-4/3)=(-4/3)sqrt(2-4/3)=(-4/3)sqrt(2/3)=-1.089
c) Inflection point
Let's find the second derivative of the function by using quotient rule,
y''=(1/2)(3sqrt(2+x)-(3x+4)(1/2)(2+x)^(-1/2))/(2+x)
y''=(1/2)(3sqrt(2+x)-(3x+4)/(2sqrt(2+x)))/(2+x)
y''=(1/4)(6(2+x)-(3x+4))/(2+x)^(3/2)
y''=(3x+8)/(4(2+x)^(3/2))
Let's find inflection point by solving x for y''=0,
(3x+8)/(4(2+x)^(3/2))=0
3x+8=0 ,harrx=-8/3
However x=-8/3 is not in the domain of the function , so there are no inflection points.