已知二次函数fx＝ax2+bx+c,若在|x|≤1时,|fx|≤1,求证：当|x|≤1时,|2a+b|≤4

根据已知可得 |f(-1)| ≤ 1,|f(0)| ≤ 1,|f(1)| ≤ 1 ,
也即 |a-b+c| ≤ 1,|c| ≤ 1,|a+b+c| ≤ 1 ,
由于 |2a+b| = |3/2*(a+b+c)+1/2*(a-b+c)-2c|
≤ 3/2*|a+b+c|+1/2*|a-b+c|+2|c| ≤ 3/2+1/2+2=4 ,
且 |-2a+b| = | -1/2*(a+b+c)-3/2*(a-b+c)+2c|
≤ 1/2*|a+b+c|+3/2*|a-b+c|+2|c| ≤ 1/2+3/2+2=4 ,
而 y = 2ax+b 是直线段,
所以,当 |x| ≤ 1 时,|2ax+b| ≤ 4 .