Answer
The domain of $t^{-1}(y)$ is all real numbers, hence the range of $t(a)$ is all real numbers.
Work Step by Step
Since we can take the cube root of any number, the domain of $t(a)$ is all real numbers. Since the range of $t(a)$ is the domain of its inverse function, we first compute the inverse of $t(a)$. Let $t(a)=y$. Solving for $a$ gives
$$
\begin{aligned}
y & =\sqrt[3]{a+1} \\
y^3 & =a+1 \\
a & =y^3-1 \\
t^{-1}(y) & =y^3-1
\end{aligned}
$$
Since $y^3-1$ is defined for any $y$, the domain of $t^{-1}(y)$ is all real numbers, hence the range of $t(a)$ is all real numbers.
