585. $y=x^{6}$.
Solution. $1^{0} . y^{\prime}=6 x^{5}$.

$2^{0} .6 x^{5}=0$, i.e. $x=0$.

$3^{0} \cdot y^{\prime \prime}=30 x^{4}$.

$4^{0}$. Since $y^{\prime \prime}(0)=0$, the method of investigation using the second derivative is not applicable.

We will investigate the function for an extremum using the first derivative. If $x<0$, then $y^{\prime}<0$; if $x>0$, then $y^{\prime}>0$. Therefore, at the point $x=0$ the function has a minimum $y=0$.