What is the domain of g(s)= 1/(sqrt(s^2-4-2)) ?

The domain of g(s)= 1/(sqrt(s^2-4-2)) is s sqrt(6)

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domain of g(s)= 1/(sqrt(s^2-4-2))

domain

g (s) = \frac{1}{s ^{2} - 4 - 2}

s < - 6 or s > 6

Interval Notation

(- \infty, - 6) \cup (6, \infty)

Solution steps

Find non-negative values for radicals:

s \leq - 6 or s \geq 6

Find undefined (singularity) points:

s = 6, s = - 6

Combine real regions and undefined points for final function domain

s < - 6 or s > 6

f (x) = x^{\frac{3}{4}}

f (x) = \frac{5 x - 2}{x - 4}

2 x^{3} + 3 x^{2} - 12 x - 18

f (x) = \frac{4 - t ^{2}}{2 - t}

f (x) = - \frac{3}{2} x^{2} + 44.5 x

What is the domain of g(s)= 1/(sqrt(s^2-4-2)) ?
The domain of g(s)= 1/(sqrt(s^2-4-2)) is s<-sqrt(6)\lor s>sqrt(6)