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

The domain of f(x)=sqrt(1/(x^2)-2) is -1/(sqrt(2))<= x<0\lor 0<x<= 1/(sqrt(2))

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

domain

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

- \frac{1}{2} \leq x < 0 or 0 < x \leq \frac{1}{2}

Interval Notation

[- \frac{1}{2}, 0) \cup (0, \frac{1}{2}]

Solution steps

Find non-negative values for radicals:

- \frac{1}{2} \leq x < 0 or 0 < x \leq \frac{1}{2}

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

- \frac{1}{2} \leq x < 0 or 0 < x \leq \frac{1}{2}

\frac{5}{9 + x}

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

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

\frac{x}{- 8 x + 81}

- 4 x + 2

What is the domain of f(x)=sqrt(1/(x^2)-2) ?
The domain of f(x)=sqrt(1/(x^2)-2) is -1/(sqrt(2))<= x<0\lor 0<x<= 1/(sqrt(2))