What is the domain of f(x)= 1/(sqrt(6-2x-x^2)) ?

The domain of f(x)= 1/(sqrt(6-2x-x^2)) is -1-sqrt(7)<x<sqrt(7)-1

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domain of f(x)= 1/(sqrt(6-2x-x^2))

domain

f (x) = \frac{1}{6 - 2 x - x ^{2}}

- 1 - 7 < x < 7 - 1

Interval Notation

(- 1 - 7, 7 - 1)

Solution steps

Find non-negative values for radicals:

- 7 - 1 \leq x \leq 7 - 1

Find undefined (singularity) points:

x = - 1 - 7, x = 7 - 1

Combine real regions and undefined points for final function domain

- 1 - 7 < x < 7 - 1

f (x) = (x - 2)^{3} (3 x + 14)

lo g_{3} (x - 3)

\frac{5}{x - 3}

f (x) = \frac{3 x - 1}{x - 5}

f (x) = x - 4 + 9 - x

What is the domain of f(x)= 1/(sqrt(6-2x-x^2)) ?
The domain of f(x)= 1/(sqrt(6-2x-x^2)) is -1-sqrt(7)<x<sqrt(7)-1