Question

find domain of f(x) =√2-2x-xsquare

Answer 1

Hope u like my process
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$= > f(x) = \sqrt{2 - 2x - {x}^{2} } \\ \\ or. \: \: f(x) = \sqrt{3 - 1 - 2x - {x}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{3 - ( {x}^{2} + 2x + 1)} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ {( {3} )} - {(x +1 )}^{2} }$
Now to get the domain f(x) ≠imaginary.

i.e.

=> (x+1)² should not be greater than 3

So,

=> x + 1 = ± (3)^½

$or. \: \: x = ( \sqrt{3} - 1)...or..( - 1 - \sqrt{3} )$

Thus x should not exceed them.
So,

$= > - 1 - \sqrt{3} \leqslant x \leqslant \sqrt{3 } - 1$
So the domain lies at
(-1-sqrt (3), 0)U(0,sqrt (3) - 1)

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