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

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gsarora17 | (Level 2) Associate Educator

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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`



Now evaluate `lim_(x->-oo)ln(x^2+1)/x`

Apply L'Hospital rule, Test condition:`oo/(-oo)`




Apply L'Hospital rule,Test L'Hospital condition:`(-oo)/oo`



`=lim_(x->-oo)1/x =0`


Find b by computing `lim_(x->-oo)f(x)-mx`



apply the limit chain rule



Since the result is not a finite constant ,so there is no horizontal asymptote at


Now let's find asymptote for `x->oo`

Find m by computing`lim_(x->oo)f(x)/x`


Let's find `lim_(x->oo)ln(x^2+1)/x`

apply L'Hospital rule, Test L'Hospital condition:`oo/oo`




apply L'Hospital rule , Test L'Hospital condition:





Find b by computing `lim_(x->oo)f(x)-mx`


Apply limit chain rule


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,




Find critical numbers by setting f'(x)=0





Now check the sign of f'(x) in the intervals (-`oo` ,-1) and (-1,`oo` ) by plugging test values -2 and 0 respectively,




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,



Set f''(x)=0 and solve for x,



`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` )





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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