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

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Chapter 4, Review - Problem 22 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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

a) Asymptotes

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

`1-x^2=0rArr(1+x)(1-x)=0`

`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

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

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

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

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

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

`x^2=-1rArrx=+-i`

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` )

`y'(-2)=(1+(-2)^2)/(1-(-2)^2)^2=5/9`

`y'(0.5)=(1+(0.5)^2)/(1-0.5^2)^2=2.22`

`y'(2)=(1+2^2)/(1-2^2)^2=5/9`

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

c) Inflection points

`y''=((1-x^2)^2(2x)-(1+x^2)(2)(1-x^2)(-2x))/(1-x^2)^4`

`y''=((1-x^2)((1-x^2)2x+4x(1+x^2)))/(1-x^2)^4`

`y''=(2x-2x^3+4x+4x^3)/(1-x^2)^3`

`y''=(2x^3+6x)/(1-x^2)^3`

`y''=(2x(x^2+3))/(1-x^2)^3`

For inflection points y''=0

`(2x(x^2+3))/(1-x^2)^3=0`

`2x(x^2+3)=0`

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

ignore the points that are complex,

So , Inflection point is at x=0

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