Question

求y=log1/3(1-1/2sinx)的值域

Answer 1

从定义域求值域

2sinx+1>0

y=log(2sinx+1)

=log(2sinx+1)/log(1/2)

=-log(2sinx+1)

-y=log(2sinx+1)

2^(-y)=2sinx+1

2^(-y)>0

(1/2)^y>0

y取任意实数,(1/2)^y>0

1>=sinx,

3>=2sinx+1

2^(-y)=log3

y值域>=log3 [log3 ,+∞)