Question

求函数f(x)=4x+1+根号1-2x的值域

Answer 1

令√(1-2x)=t>=0

则x=(1-t^2)/2

f(x)=2-2t^2+t=2-2(t^2-1/2)=17/8-2(t-1/4)^2

当t=1/4时,fmax=17/8

所以值域为f(x)