What is the domain of y=sqrt((\sqrt{25-4x^2))+4} ?

The domain of y=sqrt((\sqrt{25-4x^2))+4} is -5/2 <= x<= 5/2

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domain of y=sqrt((\sqrt{25-4x^2))+4}

domain

y = (25 - 4 x^{2}) + 4

- \frac{5}{2} \leq x \leq \frac{5}{2}

Interval Notation

[- \frac{5}{2}, \frac{5}{2}]

Solution steps

Find non-negative values for radicals:

- \frac{5}{2} \leq x \leq \frac{5}{2}

The function domain

- \frac{5}{2} \leq x \leq \frac{5}{2}

\frac{9}{3 x - 1}

10 x + 3

y = \frac{x}{4 - x ^{2}}

f (x) = \frac{- 2 x + 4}{x - 6}

5 x - 15

What is the domain of y=sqrt((\sqrt{25-4x^2))+4} ?
The domain of y=sqrt((\sqrt{25-4x^2))+4} is -5/2 <= x<= 5/2