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

### Textbook Question

Chapter 4, Review - Problem 26 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=sqrt(1-x)+sqrt(1+x)`

Domain of function: `-1<=x<=1`

a) Asymptotes

a) Asymptotes

The function has no undefined points, so there are no vertical asymptotes.

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

However `+-oo` is not in the domain , so there are horizontal asymptotes.

b) Maxima/Minima

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

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

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

`1/2(1/sqrt(1+x)-1/sqrt(1-x))=0`

`1/sqrt(1+x)=1/sqrt(1-x) rArrsqrt(1+x)=sqrt(1-x)`

squaring both the sides yields,

`1+x=1-xrArr2x=0`

`x=0`

Now let's find Maxima/Minima ; which shall be at either critical numbers or at the end points of the function.

`y(0)=sqrt(1-0)+sqrt(1+0)=2`

`y(-1)=sqrt(1-(-1))+sqrt(1-1)=sqrt(2)`

`y(1)=sqrt(1-1)+sqrt(1+1)=sqrt(2)`

Global Maximum=2 at x=0

Global Minimum = `sqrt(2)` at x=-1 and x=1

c) Inflection Points

Let's find the second derivative

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

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

For inflection points, solve for x at y''=0,

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

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

squaring both the sides and refining,

`(1+x)^3=(1-x)^3`

`1+x^3+3x(1+x)=1-x^3-3x(1-x)`

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

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

x=0 , ignore complex zeros

However if we plug in x=0 in y'' it is not true.

so there is no solution of x, hence there are no inflection points.

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