What is the domain of f(x)=\sqrt[3]{x^2+1} ?

The domain of f(x)=\sqrt[3]{x^2+1} is -infinity <x<infinity

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domain of f(x)=\sqrt[3]{x^2+1}

domain

f (x) = 3 x^{2} + 1

- \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

2 - \frac{x}{2}

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

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

f (x) = \frac{4 x + 5}{x ^{2} - 7}

f (x) = x^{4} + 5

What is the domain of f(x)=\sqrt[3]{x^2+1} ?
The domain of f(x)=\sqrt[3]{x^2+1} is -infinity <x<infinity