We need to determine the condition under which $\sqrt[n]{x}$ is never negative when $n$ is an even natural number.

1. Problem Description

We need to determine the condition under which $\sqrt[n]{x}$ is never negative when $n$ is an even natural number.

2. Solution Steps

When $n$ is an even natural number, $\sqrt[n]{x}$ is defined only for non-negative values of $x$.

That is, $x \ge 0$.

When $x=0$, then $\sqrt[n]{0}=0$, which is non-negative.

When $x>0$, $\sqrt[n]{x}$ will always be a positive number.

For example, if $n=2$ and $x=4$, $\sqrt[2]{4} = 2$, which is positive.

If $n=4$ and $x=16$, $\sqrt[4]{16} = 2$, which is positive.

Therefore, $\sqrt[n]{x}$ is never negative when $x \ge 0$ and $n$ is an even natural number.

3. Final Answer

$x \ge 0$