f′(x)=1/[2√(1+x)]-1/[2(1-x)],令f′(x)=0,则x=0.
①因为f(x)的定义域为-1≤x≤1,故函数f(x)的单调区间为[-1,0)和(0,1],且-1≤x<0时函数f(x)为单调减,0<x≤1时函数f(x)为单调增.
②稍候.
f′(x)=1/[2√(1+x)]-1/[2(1-x)],令f′(x)=0,则x=0.
①因为f(x)的定义域为-1≤x≤1,故函数f(x)的单调区间为[-1,0)和(0,1],且-1≤x<0时函数f(x)为单调减,0<x≤1时函数f(x)为单调增.
②稍候.