作业帮二次函数f（x）=ax2+bx+c在[0.1]上的值的绝对值不超过1,试问│a│+│b│+│c│的最大可能值是多少?首先f(0)=∣c∣,f(1)=∣a+b+c∣,f(1/2)=∣a/4+b/2+c∣≤1 于是∣b∣=∣4(a/4+b/2+c)-(a+b+c)-3c∣≤∣4(a/4+b/2+c)
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![二次函数f（x）=ax2+bx+c在[0.1]上的值的绝对值不超过1,试问│a│+│b│+│c│的最大可能值是多少?首先f(0)=∣c∣,f(1)=∣a+b+c∣,f(1/2)=∣a/4+b/2+c∣≤1 于是∣b∣=∣4(a/4+b/2+c)-(a+b+c)-3c∣≤∣4(a/4+b/2+c)](/uploads/image/z/15083766-54-6.jpg?t=%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dax2%2Bbx%2Bc%E5%9C%A8%5B0.1%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E4%B8%8D%E8%B6%85%E8%BF%871%2C%E8%AF%95%E9%97%AE%E2%94%82a%E2%94%82%2B%E2%94%82b%E2%94%82%2B%E2%94%82c%E2%94%82%E7%9A%84%E6%9C%80%E5%A4%A7%E5%8F%AF%E8%83%BD%E5%80%BC%E6%98%AF%E5%A4%9A%E5%B0%91%3F%E9%A6%96%E5%85%88f%280%29%3D%E2%88%A3c%E2%88%A3%2Cf%281%29%3D%E2%88%A3a%2Bb%2Bc%E2%88%A3%2Cf%281%2F2%29%3D%E2%88%A3a%2F4%2Bb%2F2%2Bc%E2%88%A3%E2%89%A41+%E4%BA%8E%E6%98%AF%E2%88%A3b%E2%88%A3%3D%E2%88%A34%28a%2F4%2Bb%2F2%2Bc%29-%28a%2Bb%2Bc%29-3c%E2%88%A3%E2%89%A4%E2%88%A34%28a%2F4%2Bb%2F2%2Bc%29)
二次函数f（x）=ax2+bx+c在[0.1]上的值的绝对值不超过1,试问│a│+│b│+│c│的最大可能值是多少?首先f(0)=∣c∣,f(1)=∣a+b+c∣,f(1/2)=∣a/4+b/2+c∣≤1 于是∣b∣=∣4(a/4+b/2+c)-(a+b+c)-3c∣≤∣4(a/4+b/2+c)
二次函数f（x）=ax2+bx+c在[0.1]上的值的绝对值不超过1,试问│a│+│b│+│c│的最大可能值是多少?
首先f(0)=∣c∣,f(1)=∣a+b+c∣,f(1/2)=∣a/4+b/2+c∣≤1
于是∣b∣=∣4(a/4+b/2+c)-(a+b+c)-3c∣≤∣4(a/4+b/2+c)∣+∣(a+b+c)∣+∣3c∣≤8
∣a∣=∣4(a/4+b/2+c)-2(a+b+c)-2c∣≤ ∣4(a/4+b/2+c)∣+∣2(a+b+c)∣+∣2c∣≤8
│a│+│b│+│c│≤8+8+1=17
我已经知道答案,但不懂每一步的原因,
又当a=8,b=-8,c=1时,f(x)=8x^2-8x+1∈[-1,1],所以│a│+│b│+│c│的最大可能值为17
二次函数f（x）=ax2+bx+c在[0.1]上的值的绝对值不超过1,试问│a│+│b│+│c│的最大可能值是多少?首先f(0)=∣c∣,f(1)=∣a+b+c∣,f(1/2)=∣a/4+b/2+c∣≤1 于是∣b∣=∣4(a/4+b/2+c)-(a+b+c)-3c∣≤∣4(a/4+b/2+c)
由已知|f(1)| = | a + b +c | ≤ 1
|f(-1)| = | a - b +c | ≤ 1
两式相加有 | a + b +c | + | a - b +c | ≤ 2
再由三角不等式有 ：2| a + c | = | (a + b +c) + (a - b +c) |
≤ | a + b +c | + | a - b +c | ≤ 2
得 | a + c | ≤ 1
另一方面 | f(0) | =| c | ≤ 1
同样由三角不等式有 |a| =|a + c - c| ≤ |a+c|+|-c| ≤ 2