作业帮已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时a=1时,求证h(x)在x属于（1,+∞）上单调递增,并证明函数h(x)有两个零点；a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x求导得x∈(1,
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 13:12:44
![已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时a=1时,求证h(x)在x属于（1,+∞）上单调递增,并证明函数h(x)有两个零点；a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x求导得x∈(1,](/uploads/image/z/5191750-46-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog2%28%28x-1%29%2F%28x%2B1%29%29%2Cg%28x%29%3D2ax%2B1-a%2C%E5%8F%88h%28x%29%3Df%28x%29%2Bg%28x%29a%3D1%E6%97%B6a%3D1%E6%97%B6%2C%E6%B1%82%E8%AF%81h%28x%29%E5%9C%A8x%E5%B1%9E%E4%BA%8E%EF%BC%881%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E5%B9%B6%E8%AF%81%E6%98%8E%E5%87%BD%E6%95%B0h%28x%29%E6%9C%89%E4%B8%A4%E4%B8%AA%E9%9B%B6%E7%82%B9%EF%BC%9Ba%3D1%E6%97%B6%2Ch%28x%29%3Df%28x%29%2Bg%28x%29%3Dlog%282%29%5B%28x-1%29%2F%28x%2B1%29%5D%2B2x%3Dlog%282%29%28x-1%29-log%282%29%28x%2B1%29%2B2x%E6%B1%82%E5%AF%BC%E5%BE%97x%E2%88%88%281%2C)
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时a=1时,求证h(x)在x属于（1,+∞）上单调递增,并证明函数h(x)有两个零点；a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x求导得x∈(1,
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时
a=1时,求证h(x)在x属于（1,+∞）上单调递增,并证明函数h(x)有两个零点；
a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x
求导得x∈(1,+∞),∴x^2-1>0,∴h'(x)>0,h(x)在(1,+∞)上为单调递增函数
∵f(1)->-∞,∴h(1)->-∞,又h(x)在(1,+∞)上为单调增函数,
∴h(x)在(1,+∞)上必有且仅有一个零点
又当a=1时,h(x)为奇函数,由奇函数的对称性可知,h(x)在(-∞,-1)上必为单调增函数
∴h(x)在(-∞,-1)上必有且仅有一个零点 ∴函数h(x)有两个零点
为什么h'(x)=1/[(x-1)ln2]-1/[(x+1)ln2]+2=1/ln2*[(x+1-x+1)/(x^2-1)]+2=2/[ln2*(x^2-1)]+2
怎么求?
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时a=1时,求证h(x)在x属于（1,+∞）上单调递增,并证明函数h(x)有两个零点；a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x求导得x∈(1,
第一步求导,用对数求导公式：[log(a)x]'=a^x*lna
第二步就是把1/ln2提出来,然后里面通分啦