Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 50

a) Let us check every undefined point a and check if the following is true:
lim_(x->a^+)f(x)=+-oo
lim_(x->a^-)f(x)=+-oo
f(x) is not defined at x=0
Check for vertical asymptote at x=0
:.lim_(x->0^-)(x-1/6x^2-2/3ln(x)) undefined
lim_(x->0^+)(x-1/6x^2-2/3ln(x))
=lim_(x->0^+)(x)-lim_(x->0^+)(1/6x^2)-lim_(x->0^+)(2/3ln(x))
=0-0-(-oo)=oo
The vertical asymptotes are x=0
Horizontal Asymptotes
Check if at x->+-oo , the function behaves as a line y=mx+b
-oo is not in the domain , so no horizontal asymptote at -
Find an asymptote for x->oo
Compute lim_(x->oo)f(x)/x to find m
lim_(x->oo)(x-1/6x^2-2/3ln(x))/x=-oo
The slope is not a finite constant, therefore no horizontal asymptote atoo
b)
f'(x)=1-1/6(2x)-2/3(1/x)
f'(x)=1-1/3x-2/(3x)
Find critical points by setting f'(x)=0
1-1/3x-2/(3x)=0
3x-x^2-2=0
x^2-3x+2=0
(x-2)(x-1)=0
x=2 , x=1
Check the sign of f'(x) by plugging test point in the intervals (0,1) , (1,2) and (2,oo )
f'(1/2)=1-1/3(1/2)-2/(3*1/2)=-1/2
f'(3/2)=1-1/3(3/2)-2/(3*3/2)=1/18
f'(6)=1-1/3(6)-2/3*6=-10/9
Since f'(1/2) is <0 , function is decreasing in the interval (0,1)
f'(3/2) is>0 , so function is increasing in the interval (1,2)
f'(6) is <0 , so function is decreasing in the interval (2,oo )
So the Local maximum is at x=2
f(2)=1-1/3(2^2)-2/3ln(2)=4/3-2/3ln(2)
Local minimum is at x=1
f(1)=1-1/6(1^2)-2/3ln(1)=5/6
c)
f''(x)=-1/3-2/3(-1)x^-2
f''(x)=-1/3+2/(3x^2)
set f''(x)=0
2/3x^2=1/3
x=+-sqrt(2)
reject -sqrt(2) as it is not in the domain
Let us check f''(x) for concavity by plugging in the test points in the intervals (0,sqrt(2) ) and (sqrt(2) ,oo )
f''(1)=-1/3+2/(3*1^2)=1/3
f''(2)=-1/3+2/(3*2^2)=-1/6
f''(1)>0 , so function is concave up in the interval (0,sqrt(2) )
f''(2)<0 , so function is concave down in the interval(sqrt(2) ,oo )
Since there is change in concavity, so Inflection point is at x=sqrt(2)