Question

已知函数g(x)的值域为[八分之三,九分之四],求函数f(x)=g(x)+根号下1-2g(x)的值域

Answer 1

设t=√[1-2g(x)] ,

由于g(x)属于[3/8,4/9]

所以t属于[1/3,1/2]

且g(x)=(1-t^2)/2

故

f(x)=(1-t^2)/2 +t

=-(1/2)t^2+t+1/2

=-(1/2)(t^2-2t+1-2)

=(-1/2)(t-1)^2+1

由于t属于[1/3,1/2]

故f(x)属于[7/9,7/8]