Question

Найдите наибольшее значение функции
y=3−log(2)(x^2−6x+11)

Answer 1

$f(x)=3-log_2(x^2-6x+11)\\f'(x)=0-(log_2(x^2-6x+11))'*(x^2-6x+11)'=-\frac{2x-6}{ln(2)(x^2-6x+11)}=\\\\=\frac{6-2x}{ln(2)(x^2-6x+11)} \\\\f'(x)=0\\\\6-2x=0\\x^2-6x+11\neq 0\\\\x=3\\D<0=>x\notin R\\\\+++++[3]-----\\\\x_{max}=3=>f_{max}(x)=f(3)=3-log_2(3^2-6*3+11)=3-log_2(2)=3-1=2\\\\f_{max}(x)=2$