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

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

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

domain

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

- 1 < x \leq 1

Interval Notation

(- 1, 1]

Solution steps

Find non-negative values for radicals:

- 1 \leq x \leq 1

Check exponents domain:

- 1 < x \leq 1

Combine real regions and undefined points for final function domain

- 1 < x \leq 1

f (x) = \frac{2 x + 5}{x - 3}

x (x + 3)^{\frac{2}{5}}

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

f (x) = 9 x - x^{3}

f (x) = \frac{x ^{2}}{4 - x ^{2}}

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