find the domain of:

Question

f(x)= √( 1-√( 1-x²))   and f(x)= √( 1- x²)+√( x²-1)

Answer 1

f(x)= √( 1-√( 1-x²)) has an inner square root, so |x| ≤1. This guarantees that the computation will always work.

 

For the second function (also unfortunately also named f), you get }x| ≤ 1 for the first radical to work, and |x| ≥ 1 for the second radical; therefore the intersection of these leaves you with the domain x=1...nothing else works.

 

So f(x) = 0 !!! but the domain remains as x=1 ONLY!  Great trick question that only a Teacher could love.

 

 

Answer 2

When looking for the domain of a function, ask yourself: what can't x be? Here, you cannot take the square root of a negative number. (zero works though) So set up the problem as an inequality.

 

f(x)= √( 1-√( 1-x²))  

 

1 - x² ≥ 0

1 ≥ x²

 

from this we get the interval: -1 ≤ x ≤ 1

 

This limits the inside square root: 0 ≤ √( 1-x²) ≤ 1

which satisfies the domain of the outer square root ≥ 0

 

Domain: {-1 ≤ x ≤ 1}

 

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f(x)= √( 1- x²)+√( x²-1)

 

1 - x² ≥ 0

1 ≥ x²

-1 ≤ x ≤ 1

 

x² - 1 ≥ 0

x² ≥ 1

-1 ≥ x ≥ 1

 

EDIT: The domains of each radical intersect at x = 1 and x = -1 only.