`y = x/(1 - x^2)` 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

Vertical asymptotes are the zeros of the denominator of the function.


`x=-1 , x=1`

Vertical asymptotes are x=1 and x=-1

Degree of numerator=1

Degree of denominator=2

Since degree of denominator degree of numerator,

So, Horizontal asymptote is x-axis i.e. y=0

b) Maxima/Minima




To find critical numbers , solve for x at y'=0

`(1+x^2)/(1-x^2)^2=0 rArr1+x^2=0`


Since there is no real solution , so there are no critical points.

Domain : -1`<` x`<` 1

Check the sign of y' by plugging test points in the intervals (-`oo` ,-1),(-1,1) and (1,`oo` )




There is no change in sign of y' , so there are maxima and minima.

c) Inflection points






For inflection points y''=0



`x=0 , x^2=-3`

ignore the points that are complex,

So , Inflection point is at x=0

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