To find the domain of the function $f(t) = \sqrt[3]{8t - 1}$, we need to determine the set of all possible values of $t$ for which the function is defined.

Analysis:

• The cube root function, $\sqrt[3]{x}$, is defined for all real numbers $x$. Unlike the square root function, the cube root function does not have any restrictions such as needing the argument to be non-negative.

Therefore, for the function $f(t) = \sqrt[3]{8t - 1}$, there are no restrictions on the value of $t$. The expression $8t - 1$ can take any real value, meaning that the function is defined for all real values of $t$.

Conclusion:

The domain of $f(t) = \sqrt[3]{8t - 1}$ is all real numbers.

Domain in interval notation:

$\text{Domain} = (-\infty, \infty)$

Would you like more details or have any questions?

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Here are 5 related questions for practice:

1. What is the domain of $g(x) = \sqrt{2x - 5}$?
2. Determine the domain of $h(x) = \frac{1}{x - 3}$.
3. Find the domain of $f(x) = \ln(x - 4)$.
4. What is the range of the function $f(t) = \sqrt[3]{8t - 1}$?
5. Is the function $f(x) = \sqrt{9 - x^2}$ defined for $x = 4$?

Tip: When finding the domain of a function, always consider the operations involved (e.g., square roots, logarithms, divisions) to identify any restrictions on the input values.