# What is function real, f(x)=ax^2+bx+c when modulus f(x) <= (x-1)^2? modulus f(x) is > 0

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### 1 Answer

You need to use absolute value property, such that:

`|f(x)| <= (x - 1)^2 => -(x - 1)^2 <= ax^2+bx+c <= (x - 1)^2`

Considering the left inequality, yields:

`-(x - 1)^2 <= ax^2 + bx + c =>` `-x^2 + 2x - 1 <= ax^2 + bx + c`

`ax^2 + x^2 + bx - 2x + c + 1 >= 0`

`x^2(a + 1) + x(b - 2) + c + 1 >= 0`

The inequality holds for `Delta = (b - 2)^2 - 4(a+1)(c+1) = 0` and `a + 1 > 0` such that:

`(b - 2)^2 = 4(a+1)(c+1)`

`a + 1 > 0 => a > -1 => a in (-1,oo)`

Considering `a = 1, b = -2, c = 1` yields `f(x) = 2x^2 - 4x + 2`

Factoring out 2 yields:

`f(x) = 2(x^2 - 2x + 1) => f(x) = 2(x - 1)^2`

Considering the right inequality, yields:

`ax^2 + bx + c <= (x - 1)^2`

`ax^2 + bx + c - (x - 1)^2 <= 0`

`ax^2 - x^2 + bx + 2x + c - 1 <= 0`

`x^2(a - 1) + x(b + 2) + c - 1 <= 0`

The inequality holds for `Delta = (b + 2)^2 - 4(a-1)(c-1) = 0` and `a - 1 < 0` such that:

`(b + 2)^2 = 4(a-1)(c-1)`

`a < 1 => a in (-oo,1)`

Considering `a = -1, b = 2, c = -1` yields `f(x) = -2x^2 + 4x - 2`

Factoring out 2 yields:

`f(x) = -2(x^2 - 2x + 1) => f(x) = -2(x - 1)^2`

**Hence, evaluating the functions f(x), using the condition provided by the problem `|f(x)| <= (x - 1)^2` yields **`f(x) = +-2(x - 1)^2.`