What is the domain of f(x)= 1/(sqrt(2-2x-x^2)) ?

The domain of f(x)= 1/(sqrt(2-2x-x^2)) is -1-sqrt(3)<x<sqrt(3)-1

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domain of f(x)= 1/(sqrt(2-2x-x^2))

domain

f (x) = \frac{1}{2 - 2 x - x ^{2}}

- 1 - 3 < x < 3 - 1

Interval Notation

(- 1 - 3, 3 - 1)

Solution steps

Find non-negative values for radicals:

- 3 - 1 \leq x \leq 3 - 1

Find undefined (singularity) points:

x = - 1 - 3, x = 3 - 1

Combine real regions and undefined points for final function domain

- 1 - 3 < x < 3 - 1

- \frac{1}{2 ( 5 + x ) ^{\frac{3}{2}}}

y = (x + 4) - 6

f (x) = \frac{9}{x - 7}

u (x) = 2 - x

+ 20 x - 4

What is the domain of f(x)= 1/(sqrt(2-2x-x^2)) ?
The domain of f(x)= 1/(sqrt(2-2x-x^2)) is -1-sqrt(3)<x<sqrt(3)-1