Answer
Domain: $a\geq 0$
Range: $y\geq 0$
Work Step by Step
Given
\begin{equation}
k(a)=3 \sqrt[4]{a}.
\end{equation} The radicand of an even radical function must be positive. This means that we must have $a\geq 0$. Also, we see that the function's value $y=k(a)$ must satisfy $y\geq 0$ because $$\sqrt[4]a\geq 0\implies 3\sqrt[4]a\geq 0\implies y\geq 0.$$ We got: \begin{equation}
\begin{aligned}
\textbf{Domain :}\quad & a\geq 0\\
\textbf{Range :}\quad & y\geq 0\\\\
\end{aligned}
\end{equation}
