What is the domain of y=sqrt(1-(x^2)/9) ?

The domain of y=sqrt(1-(x^2)/9) is -3<= x<= 3

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domain of y=sqrt(1-(x^2)/9)

domain

y = 1 - \frac{x ^{2}}{9}

- 3 \leq x \leq 3

Interval Notation

[- 3, 3]

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

The function domain

- 3 \leq x \leq 3

\frac{1}{3} (x - 2.1)^{2} + 7.9

ln (- \frac{- 6 - 3 x ^{2}}{x ^{2} + 1})

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

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

f (x) = \frac{x - 9}{x - 2}

What is the domain of y=sqrt(1-(x^2)/9) ?
The domain of y=sqrt(1-(x^2)/9) is -3<= x<= 3