Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 26

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.