Найдите точку минимума функции y=(2x^3-3x^2+1)/8x^3

$y= \frac{2x^3-3x^2+1}{8x^3} \\ \\ y'= \frac{(2x^3-3x^2+1)'*8x^3-(8x^3)'*(2x^3-3x^2+1)}{64x^6} = \\ = \frac{(6x^2-6x)*8x^3-24x^2(2x^3-3x^2+1)}{64x^6} = \frac{48x^5-48x^4-48x^5+72x^4-24x^2}{64x^6} = \\ = \frac{24x^4-24x^2}{64x^6} = \frac{24x^2(x^2-1)}{64x^6} = \frac{3(x^2-1)}{8x^4} \\ \\ \frac{3(x^2-1)}{8x^4}=0 \\ \\ x^2-1=0 \\ x^2=1 \\ x=б1 \\ \\ 8x^4 \neq 0 \\ x \neq 0$

_______+______ $-1$ ______-______ $1$ _______+_______

Ответ: $x_{min}=1$