`y = x + ln(x^2 + 1)` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

Textbook Question

Chapter 4, Review - Problem 34 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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`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 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)`

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)

 

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