567. $y=(x-5) e^{x}$.
Solution. $1^{0} . y^{\prime}=(x-5)^{\prime} e^{x}+$ $+\left(e^{x}\right)^{\prime}(x-5)=e^{x}+e^{x}(x-5)=e^{x}(x-4)$. $2^{0} . e^{x}(x-4)=0 ; e^{x} \neq 0 ; x-4=0 ; x=4$.

$3^{\prime \prime}$. In the interval $(-\infty, 4)$ the derivative is negative, and in the interval $(4, \infty)$ it is positive: Therefore, at $x=4$ the function has a minimum.

$4^{0} . f(4)=y_{\min }=-e^{4}$.