Question

設(shè)f(x)=ax²+bx+c,當(dāng)|x|≤1時,總有|f(x)|≤1,求證:當(dāng)|x|≤2時,|f(x)|≤7.

Answer 1

證明:由于f(x)是二次函數(shù),|f(x)|在［-2,2］上的最大值只能是|f(2)|,|f(-2)|或|f()|,故只要證明|f(2)|≤7,|f(-2)|≤7;當(dāng)||≤2時,有|f()|≤7.

由題意有|f(0)|≤1,|f(-1)|≤1,|f(1)|≤1.

由

∴|f(2)|=|4a+2b+c|=|3f(1)+f(-1)-3f(0)|≤3|f(1)|+|f(-1)|+3|f(0)|≤3+1+3=7,

|f(-2)|=|4a-2b+c|=|f(1)+3f(-1)-3f(0)|≤|f(1)|+3|f(-1)|+3|f(0)|≤1+3+3=7.

∵|b|=|f(1)-f(-1)|≤(|f(1)|+|f(-1)|)≤(1+1)=1,

∴當(dāng)||≤2時,|f()|=||=|c|=|c·|≤|c|+||·1+2×=2＜7.

因此當(dāng)|x|≤2時,|f(x)|≤7.