Question

Найти min функции y=3x4-4x3+5 на интервале [-2;3]

Answer 1

Объяснение:

$y=3x^4-4x^3+5\ \ \ \ \ [-2;3].\\y'=(3x^4-4x^3+5)'=3*4*x^3-4*3*x^2=12x^3-12x^2=0\\12x^2(x-1)=0\ |:12\\x^2*(x-1)=0\\x^2=0\\x_1=0\in.\\x-1=0\\x_2=1\in.\\y(-2)=3*(-2)^4-4*(-2)^3+5=3*16-4*(-8)+5=48+32+5=85.\\y(0)=3*0^4-4*0^3+5=5.\\y(1)=3*1^4-4*1^3+5=3-4+5=4.\\y(3)=3*3^4-4*3^3+5=3*81-4*27+5=243-108+5=140.\ \ \ \ \ \Rightarrow$

Ответ: ymin=y(1)=4.