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已知a,b,c是实数,函数f(x)=ax^2+bx+c,g(x)=ax+b,当-1<=x<=1时,/f(x)/<=1,问题在下面哦(1)求证|c|<=1
2个回答
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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
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
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