f(x)=ax^2+bx+c,g(x)=ax+b,当-1≤x≤1时,|f(x)|≤1,
(1)求证|c|≤1 (2)求证,当-1≤X≤1时,|g(X)|≤2
1.
i当a=0时,f(x)=bx+c
b≥0
-1≤x≤1
-b≤bx≤b
-b+c≤bx+c≤b+c
-b+c≤f(x)≤b+c
|f(x)|≤1,
-1≤f(x)≤1
-1≤-b+c≤f(x)≤b+c≤1
-1≤-b+c,b+c≤1
b≤1+c,b≤1-c
因b≥0
1+c≥0,1-c≥0,
c≥-1,1≥c,
-1≤c≤1
|c|≤1;
b＜0
ii当a≠0时,