What is the domain of f(x)=\sqrt[4]{25-x^2} ?

The domain of f(x)=\sqrt[4]{25-x^2} is -5<= x<= 5

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domain of f(x)=\sqrt[4]{25-x^2}

domain

- 5 \leq x \leq 5

Interval Notation

[- 5, 5]

Solution steps

Find non-negative values for radicals:

- 5 \leq x \leq 5

The function domain

- 5 \leq x \leq 5

f (x) = - 36, 0 \leq x \leq 360

x^{2} - 4 + x^{2} + 5 x + 6

\frac{x + 1}{x ^{2} - 2 x + 1}

f (t) = 4 - t^{2}

x^{2} - 4 x + 3

What is the domain of f(x)=\sqrt[4]{25-x^2} ?
The domain of f(x)=\sqrt[4]{25-x^2} is -5<= x<= 5