Question

Find the domain of the function $f\left(t\right)=\sqrt[3]{t-1}$.

Short answer

The domain of the function is $t\in \left(-\infty ,\infty \right)$.

Step-by-step solution

Step 1. By definition of domain of a function.

The given function is $f\left(t\right)=\sqrt[3]{t-1}$. Now, by definition of domain of function, the domain of the function is defined for all real values of $\left(t\right)$for which the function is defined.

Step 2. By definition of domain of cube root function.

From the cube root function $f\left(t\right)=\sqrt[3]{t-1}$, the domain is defined for all real numbers $\left(t\right)$for which the function role="math" localid="1644581918901" $f\left(t\right)=\sqrt[3]{t-1}$is defined. Since the function role="math" localid="1644581925040" $f\left(t\right)=\sqrt[3]{t-1}$is a cubic function, therefore the domain of function is defined for all real numbers.

Hence the domain of the function is $t\in \left(-\infty ,\infty \right)$.

Step 3. DESCRIPTION OF STEPS.

The domain of the function is $t\in \left(-\infty ,\infty \right)$.