已知f(x)=xlog(2)x+(1-x)log(2)(1-x)且1>x>0求f(x)的最小值 怎么解

f(x)=xlog(2)x+(1-x)log(2)(1-x)
求导数
f'(x)
=log(2)x+x/(xln2)-log(2)(1-x)-(1-x)/[(1-x)ln2]
=log(2)x+1/ln2-log(2)(1-x)-1/ln2
=log(2)x-log(2)(1-x)
=log(2)[x/(1-x)]
令f'(x)=0,即x/(1-x)=1,解得x=1/2
x<1/2时,f'(x)<0,f(x)为减函数
x>1/2时,f'(x)>0,f(x)为增函数

所以x=1/2时函数取最小值
f(1/2)=-1/2-1/2=-1