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

The domain of f(y)=sqrt(1-y^2) is -1<= y<= 1

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domain of f(y)=sqrt(1-y^2)

domain

f (y) = 1 - y^{2}

- 1 \leq y \leq 1

Interval Notation

[- 1, 1]

Solution steps

Find non-negative values for radicals:

- 1 \leq y \leq 1

The function domain

- 1 \leq y \leq 1

sin^{2} (x) + cos (x)

26 x^{5}

sin (x) + 2

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

(2 x - 5) \cdot (2 x + 5)

What is the domain of f(y)=sqrt(1-y^2) ?
The domain of f(y)=sqrt(1-y^2) is -1<= y<= 1