The correct answer is: B. The function $\sqrt[{\text{x}}]{{\text{x}}}$,(x > 0), has the global maxima at x = e
The function $\sqrt[{\text{x}}]{{\text{x}}}$,(x > 0), is a decreasing function. This means that for any two values $x_1$ and $x_2$ such that $x_1 > x_2$, we have $\sqrt[{\text{x}}]{{\text{x}}}(x_1) < \sqrt[{\text{x}}]{{\text{x}}}(x_2)$. In particular, this means that $\sqrt[{\text{x}}]{{\text{x}}}$ has no global maxima, since there is no value $x$ such that $\sqrt[{\text{x}}]{{\text{x}}}(x) > \sqrt[{\text{x}}]{{\text{x}}}(y)$ for all $y$.
The other options are all correct. The function $x^3$ has a global minimum at $x=0$, since $x^3(0)=0$ and $x^3(x)>0$ for all $x>0$. The function $|x|$ has no global minima or global maxima, since it is a non-decreasing function.