Question

найдите наименьшее значение функции g(x)=log0.5(-основание)(2-х^2)

Answer 1

 

$<var>g(x)=log_{0,5}(2-x^2), \ 2-x^2>0, x^2-2<0, (x+sqrt2)(x-sqrt2)<0, -sqrt2<x<sqrt2, \ D_g=(-sqrt2;sqrt2), \ g'(x)=(2-x^2)ln0,5cdot(2-x^2)'=-2x(2-x^2)ln0,5= \ =2x(x^2-2)ln0,5, \ g'(x)=0, 2x(x^2-2)ln0,5=0, \ x(x+sqrt2)(x-sqrt2)=0, \ x_1=0, x_2=-sqrt2notin D, x_2=sqrt2notin D, \ g(0)=log_{2^{-1}}2=-log_22=-1, \ min - (0;-1), g_{min}=-1.$