Question

Найдите наименьшее значение функции y=12x^3-18x^2+36 на отрезке [-2;2]

Answer 1

$y=12x^3-18x^2+36\; \; ,\qquad x\in [-2\, ;\, 2\; ]\\\\y'=36x^2-36x=36x(x-1)=0\; \; ,\; \; x_1=0\; ,\; \; x_2=1\\\\znaki\; y'\, :\; \; \; +++(0)---(1)+++\\\\x=-2:\; \; y(-2)=-12\cdot 8-18\cdot 4+36=-132\\\\x=0:\; \; y(0)=36\\\\x=1\, :\; \; y(1)=12\cdot 1-18\cdot 1+36=30\\\\x=2:\; \; y(2)=12\cdot 8-18\cdot 4+36=60\\\\y(naimen.)=y(-2)=-132$