What is the domain of arccos(1/(y^2)) ?

The domain of arccos(1/(y^2)) is y = 1

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domain of arccos(1/(y^2))

domain

arccos (\frac{1}{y ^{2}})

y \leq - 1 or y \geq 1

Interval Notation

(- \infty, - 1] \cup [1, \infty)

Solution steps

Find known functions domain restrictions:

y \leq - 1 or y \geq 1

Find undefined (singularity) points:

y = 0

Combine real regions and undefined points for final function domain

y \leq - 1 or y \geq 1

f (x) = 4 x^{3} + 6 x^{2} - 72 x - 9

x y^{2} + 10 = 0

lo g_{4} (x) + 2

f (x) = x^{2} + 8 x - 1

f (x) = 1 - 2^{t}

What is the domain of arccos(1/(y^2)) ?
The domain of arccos(1/(y^2)) is y<=-1\lor y>= 1