Question on domain of function

[peblez]

f( x ) = ( 1 - x^2 ) ^ 1/2

find domain

i know the answer is x <_ 1 and x >_ -1

but when i try to solve it i get different results.

this is what i did.

1 - x^2 >_ 0

x^2 < 1

x < plus or minus 1

x < 1
x < -1

So how come i get x < -1 , is it because of the negative so i have to reverse the sign of the inequality?

 

[Loren]

\(\displaystyle \sqrt{1-x^2}=0\)
\(\displaystyle 1-x^2=0\)
\(\displaystyle x^2=1\)
\(\displaystyle x=\pm1\)

Therefore, critical points are x = -1 and x = 1.

If x < -1, then f(x) is imaginary.
If -1 < x < 1. then f(x) is real.
If x < 1 then f(x) is imaginary.

 

[Dr. Flim-Flam]

f(x) = ?(1-x²) Hence, then (1-x²) ? 0, x² ? 1, Taking the square root of both side.

?x² ? ?1, |x| ? 1, By definition ?1 = 1 and ?x² = |x|, Ergo , |x| ? 1, -1 ?x ?1.

Note: Critical points are f(-1) = 0, f(1) = 0, (endpoints) and f(0) = 1 (absolute max).