帮忙解一道数学题

显示全部楼层 · 2008-9-6 15:28:40

函数f（x）＝ax+b，当｜x｜＜＝1时都有｜f（x）｜＜＝1．求证｜a｜＜＝1，｜b｜＜＝1

千问 · 2008-9-6 15:28:40

证明：因为f(x)=ax+b,f(0)=b,f(1)=a+b,f(-1)=b-a 当|x|<=1时,都有|f(x)|<=1，取x=0,则│f(0)│<=1即|b|<=1 又|f(1)|<=1,|f(-1)|<=1, a=(a+b)/2+(a-b)/2=(1/2)[f(1)-f(-1)], 所以│a│=(1/2)(│f(1)-f(-1)│)<=(1/2)(│f(1)│+│f(-1)│)<=(1/2)(1+1)=1 即|a|<=1...

千问 · 2008-9-6 15:28:40

|f（x）|＝|ax+b|<=1|ax+b|<=|a||x|+|b|<=|a|+|b|<=1所以｜a｜＜＝1，｜b｜＜＝1...