f(x)=ax^2+bx+c,g(x)=ax+b,當-1<=x<=1時，f(x)的絕對值<=1.求證:(1)當-1<=x<=1時,g(x)的絕對值<=2 (2)設a>0,當-1<=x<=1時,g(x)的最大值為2，求f(x)

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f(x)=ax^+bx+c,g(x)=ax+b,當-1≤x≤1時，|f(x)|≤1。求證:(1)當-1≤x≤1時,|g(x)|≤2 (2)設a0,當-1≤x≤1時,g(x)的最大值為2，求f(x) (1)證:|g(x)|=|ax+b|≤|ax|+|b|≤|a|+|b|①當a,b同號時:|a|+|b|=|a+b|=|(a+b+c)-c|=|f(1)-f(0)|≤|f(1)|+|f(0)|≤2∴|g(x)|≤2②當a,b異號時:|a|+|b|=|a-b|=|(a-b+c)-c|=|f(-1)-f(0)|≤|f(-1)|+|f(0)|∴|g(x)|≤2不論a,b同號或異號時都有:|g(x)|≤2(2)a0,當-1≤x≤1時,g(x)的最大值為2，①b≥0時,a,b同號?！?=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。必須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無解)∴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?！郺=2-b,c=-1∴f(x)=(2-b)x^+bx-1,(0≤b≤1)。[注:如b=0時f(x)=2x^-1且g(x)=2x,b=1時f(x)=x^+x-1且g(x)=x+1,b=1/2時f(x)=(3/2)x^+(1/2)x-1且g(x)=(3/2)x+(1/2)經檢驗都滿足要求,則f(x)不是唯一的。]②b<0時,a,b異號?！?=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必須x=-1且|f(-1)|=|f(0)|=1且a-b=2?！遼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)。。