What is the domain of f(t)=\sqrt[3]{t-1} ?

The domain of f(t)=\sqrt[3]{t-1} is -infinity <t<infinity

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domain of f(t)=\sqrt[3]{t-1}

domain

- \infty < t < \infty

Solution steps

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

- \infty < t < \infty

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

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

f (x) = \frac{x + 4 x ^{3} - 5}{5 x ^{3} - 2 x ^{2} + 2}

f (x) = \frac{6 x + 4}{x - 1}

f (x) = \frac{1}{6 x - 12}

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