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

The domain of f(x)= 1/(2x^2) is x 0

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

domain

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

x < 0 or x > 0

Interval Notation

(- \infty, 0) \cup (0, \infty)

Solution steps

Find undefined (singularity) points:

x = 0

The function domain

x < 0 or x > 0

f (x) = \frac{1}{x - 2} + 3

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

f (x) = - x^{2} + 8 x - 12

f (x) = ∣ x - 7 ∣

ln (\frac{1}{9 - x ^{2}})

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