What is the domain of f(x)=sqrt(1/(x^2)-5) ?

The domain of f(x)=sqrt(1/(x^2)-5) is -1/(sqrt(5))<= x<0\lor 0<x<= 1/(sqrt(5))

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domain of f(x)=sqrt(1/(x^2)-5)

domain

f (x) = \frac{1}{x ^{2}} - 5

- \frac{1}{5} \leq x < 0 or 0 < x \leq \frac{1}{5}

Interval Notation

[- \frac{1}{5}, 0) \cup (0, \frac{1}{5}]

Solution steps

Find non-negative values for radicals:

- \frac{1}{5} \leq x < 0 or 0 < x \leq \frac{1}{5}

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

- \frac{1}{5} \leq x < 0 or 0 < x \leq \frac{1}{5}

f (x) = x - x^{2}

f (x) = \frac{5 x}{9 x - 5}

f (x) = \frac{x + 5}{x - 6}

f (x) = - \frac{5}{4} x + 10

5 sin (\frac{1}{4} x)

What is the domain of f(x)=sqrt(1/(x^2)-5) ?
The domain of f(x)=sqrt(1/(x^2)-5) is -1/(sqrt(5))<= x<0\lor 0<x<= 1/(sqrt(5))