What is the domain of f(x)=\sqrt[3]{x^2-9} ?

The domain of f(x)=\sqrt[3]{x^2-9} is -infinity <x<infinity

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domain of f(x)=\sqrt[3]{x^2-9}

domain

- \infty < x < \infty

Interval Notation

(- \infty, \infty)

Solution steps

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

f (x) = \frac{3}{4} x + 2 - 3

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

f (x) = - cos (x)

3 \cdot e^{- 5 x}

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

What is the domain of f(x)=\sqrt[3]{x^2-9} ?
The domain of f(x)=\sqrt[3]{x^2-9} is -infinity <x<infinity