`y = root(3)(x^2 + 1)` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

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Chapter 4, Review - Problem 28 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=root(3)(x^2+1)`

a) Asymptotes

Since the function has no undefined point, so it has no vertical asymptote.

For horizontal asymptotes check if at x`->+-oo` , the function behaves as a line y=mx+b,

Compute `lim_(x->+-oo)f(x)/x` to find m

`lim_(x->+-oo)root(3)(x^2+1)/x=0`

Compute `lim_(x->+-oo)f(x)-mx` to find b,

`lim_(x->+-oo)root(3)(x^2+1)-0x=oo`

Since the result is not a finite constant , so there is no horizontal asymptote at `+-oo`

b) Maxima/Minima

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

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

Let's find critical numbers by solving x, for y'=0

`(2x)/(3(x^2+1)^(2/3))=0 ,rArrx=0`  

Let's check the sign of y' by plugging test point in the intervals (-`oo` ,0) and (0,`oo` )

`y'(-1)=-2/(3((-1)^2+1)^(2/3))=-2/(3(2)^(2/3))=-0.42`

`y'(1)=2/(3(2)^(2/3))=0.42`

There is no maximum point.

Minimum point is at x=0

c) Inflection Points

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

`y''=(2/3)(3(x^2+1)-4x^2)/(3(x^2+1)^(5/3))`

`y''=(2/9)(3-x^2)/(x^2+1)^(5/3)`

Let's find inflection points by solving x, for y''=0

`3-x^2=0 , x=+-sqrt(3)`

Inflection points are (`sqrt(3)` ,`root(3)(4)` ) and (-`sqrt(3)` ,`root(3)(4)` )

Graph: Function is plotted in red color and second derivative is plotted in green color.

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