Question

Найдите наибольшее значение функции:
y=x^4-4x^3-x^2+6x на промежутке [2;3]

Answer 1

$f(x)=x^4-4x^3-x^2+6x\\\\f'(x)=4x^3-12x^2-2x+6\\\\f'(x)=0\\\\4x^3-12x^2-2x+6=0\\\\4x^2(x-3)-2(x-3)=0\\\\2(x-3)(2x^2-1)=0\ \ \mid\div2\\\\(x-3)(2x^2-1)=0\\\\x=3\\x=\pm\frac{\sqrt{2} }{2}\ \ \notin[2;3]\\\\f(2)=2^4-4*2^3-2^2+6*2=-8\\\\f(3)=3^4-4*3^3-3^2+6*3=-18\\\\f_{max[2;3]}=-8$