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

Textbook Question

Chapter 4, Review - Problem 27 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=xsqrt(2+x)`

a) Asymptotes

Domain of function is x `>=` -2

Since the function has no undefined points , so it has no vertical asymptotes.

Horizontal Asymptotes:

Let's find the limits of the function at `+-oo`

Since -`oo` is not in the domain so there is no horizontal asymptote at -`oo`

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

`lim_(x->oo)(xsqrt(2+x))/x`

`=lim_(x->oo)sqrt(2+x)=oo`

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

b) Maxima/Minima

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

`y'=x/(2sqrt(2+x))+sqrt(2+x)`

`y'=(x+2(2+x))/(2sqrt(2+x))`

`y'=(3x+4)/(2sqrt(2+x))`

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

`(3x+4)/(2sqrt(2+x))=0`

`3x+4=0 ,=>x=-4/3`

Let's check the sign of y' by plugging test points in the intervals (-2,-4/3) and (-4/3 ,`oo` )

`y'(-1.5)=(3(-1.5)+4)/(2sqrt(2+(-1.5)))=-0.354`

`y'(0)=(3(0)+4)/(2sqrt(2+0))=4/(2sqrt(2))=sqrt(2)`

There is no maxima.

Minimum point is at x=-4/3

`y(-4/3)=(-4/3)sqrt(2-4/3)=(-4/3)sqrt(2/3)=-1.089`

c) Inflection point

Let's find the second derivative of the function by using quotient rule,

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

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

`y''=(1/4)(6(2+x)-(3x+4))/(2+x)^(3/2)`

`y''=(3x+8)/(4(2+x)^(3/2))`

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

`(3x+8)/(4(2+x)^(3/2))=0`

`3x+8=0 ,harrx=-8/3`

However x=-8/3 is not in the domain of the function , so there are no inflection points.

 

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