f(x)=ax^2+bx+c,g(x)=ax+b,当-1<=x<=1时，f(x)的绝对值<=1.求证:(1)当-1<=x<=1时,g(x)的绝对值<=2 (2)设a>0,当-1<=x<=1时,g(x)的最大值为2，求f(x)
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f(x)=ax^+bx+c,g(x)=ax+b,当-1≤x≤1时，|f(x)|≤1。求证:(1)当-1≤x≤1时,|g(x)|≤2 (2)设a0,当-1≤x≤1时,g(x)的最大值为2，求f(x) (1)证:|g(x)|=|ax+b|≤|ax|+|b|≤|a|+|b|①当a,b同号时:|a|+|b|=|a+b|=|(a+b+c)-c|=|f(1)-f(0)|≤|f(1)|+|f(0)|≤2∴|g(x)|≤2②当a,b异号时:|a|+|b|=|a-b|=|(a-b+c)-c|=|f(-1)-f(0)|≤|f(-1)|+|f(0)|∴|g(x)|≤2不论a,b同号或异号时都有:|g(x)|≤2(2)a0,当-1≤x≤1时,g(x)的最大值为2，①b≥0时,a,b同号。∵2=g(x)≤|g(x)|=|ax+b|≤|ax|+|b|≤|a|+|b|=|a+b|=|(a+b+c)-c|=|f(1)-f(0)|≤|f(1)|+|f(0)|≤2。即:2≤2∴欲使上面不等式成立。必须x=1且|f(1)|=|f(0)|=1且a+b=2∴|f(1)|=|a+b+c|=|2+c|=|f(0)|=|c|∴2+c=-c(2+c=c无解)∴c=-1∵|f(-1)|=|a-b+c|=|(a+b)-2b+c|=|2-2b-1|=|2b-1|≤1∴-1≤2b-1≤1即:0≤b≤1。∴a=2-b,c=-1∴f(x)=(2-b)x^+bx-1,(0≤b≤1)。[注:如b=0时f(x)=2x^-1且g(x)=2x,b=1时f(x)=x^+x-1且g(x)=x+1,b=1/2时f(x)=(3/2)x^+(1/2)x-1且g(x)=(3/2)x+(1/2)经检验都满足要求,则f(x)不是唯一的。]②b<0时,a,b异号。∵2=g(x)≤|g(x)|=|ax+b|≤|ax|+|b|≤|a|+|b|=|a-b|=|(a-b+c)-c|=|f(-1)-f(0)|≤|f(-1)|+|f(0)|≤2。即:2≤2必须x=-1且|f(-1)|=|f(0)|=1且a-b=2。∵|f(-1)|=|a-b+c|=|2+c|=|f(0)|=|c|=1c=-1,同理-1≤b≤0,a=b+2∴f(x)=(2+b)x^-bx-1,(-1≤b≤0)。。