`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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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



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


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

squaring both the sides yields,




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




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



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



squaring both the sides and refining,





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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