Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 34

y=f(x)=x+ln(x^2+1)
a) Asymptotes
The function has no undefined points , so it has no vertical asymptotes.
For Horozontal asymptotes , check if at x->+-oo the function behaves as a line y=mx+b
Find an asymptote for x->-oo
Find m by computing lim_(x->-oo)f(x)/x
lim_(x->-oo)(x+ln(x^2+1))/x
=lim_(x->-oo)(1+ln(x^2+1)/x)
Now evaluate lim_(x->-oo)ln(x^2+1)/x
Apply L'Hospital rule, Test condition:oo/(-oo)
lim_(x->-oo)ln(x^2+1)/x=lim_(x->-oo)((ln(x^2+1))')/(x')
=lim_(x->-oo)((2x)/(x^2+1))/1
=lim_(x->-oo)(2x)/(x^2+1)
Apply L'Hospital rule,Test L'Hospital condition:(-oo)/oo
=lim_(x->-oo)((2x)')/((x^2+1)')
=lim_(x->-oo)2/(2x)
=lim_(x->-oo)1/x =0
So,lim_(x->-oo)(x+ln(x^2+1))/x=1
Find b by computing lim_(x->-oo)f(x)-mx
lim_(x->-oo)f(x)-mx=lim_(x->-oo)(x+ln(x^2+1)-x)
=lim_(x->-oo)ln(x^2+1)
apply the limit chain rule
=lim_(x->-oo)(x^2+1)
=oo+1=oo
Since the result is not a finite constant ,so there is no horizontal asymptote at
-oo
Now let's find asymptote for x->oo
Find m by computinglim_(x->oo)f(x)/x
lim_(x->oo)(x+ln(x^2+1))/x=lim_(x->oo)(1+ln(x^2+1)/x)
Let's find lim_(x->oo)ln(x^2+1)/x
apply L'Hospital rule, Test L'Hospital condition:oo/oo
lim_(x->oo)ln(x^2+1)/x=lim_(x->oo)((ln(x^2+1))')/(x')
=lim_(x->oo)((2x)/(x^2+1))/1
=lim_(x->oo)(2x)/(x^2+1)
apply L'Hospital rule , Test L'Hospital condition:
=lim_(x->oo)2/(2x)=lim_(x->oo)1/x=0

So,lim_(x->oo)(x+ln(x^2+1))/x=1+0=1

Find b by computing lim_(x->oo)f(x)-mx
lim_(x->oo)(x+ln(x^2+1)-x)=lim_(x->oo)ln(x^2+1)
Apply limit chain rule
=lim_(x->oo)(x^2+1)=oo+1=oo
Since the result is not a finite constant , so there is no horizontal asymptote at plus infinity.
b) Maxima/Minima
Now let's find the critical numbers by taking the first derivative,
f'(x)=1+(2x)/(x^2+1)

f'(x)=(x^2+1+2x)/(x^2+1)=(x+1)^2/(x^2+1)
Find critical numbers by setting f'(x)=0
(x+1)^2/(x^2+1)=0
(x+1)^2=0
x+1=0
x=-1
Now check the sign of f'(x) in the intervals (-oo ,-1) and (-1,oo ) by plugging test values -2 and 0 respectively,

f'(-2)=(-2+1)^2/((-2)^2+1)=1/5
f'(0)=(0+1)^2/(0^2+1)=1
Since the sign of f'(x) is positive in both the intervals, so the function is increasing in both the intervals and there is no maximum/minimum.
c) Inflection points
Find the second derivative by applying quotient rule,
f''(x)=((x^2+1)2-2x(2x))/(x^2+1)^2
f''(x)=(2x^2+2-4x^2)/(x^2+1)^2=(2(1-x^2))/(x^2+1)^2
Set f''(x)=0 and solve for x,
(2(1-x^2))/(x^2+1)^2=0
1-x^2=0
x=1, x=-1
Now let's find the concavity of the curve by plugging test values in the intervals (-oo ,-1) , (-1,1) and (1,oo )
f''(-2)=(2(1-4))/((-2)^2+1)^2=-6/25

f''(0)=2
f''(2)=-6/25
So the function is concave down in the interval (-oo ,-1) and (1,oo )
function is concave up in the interval (-1,1)