MHB Solving a Third-Degree Polynomial with Real Coefficients

[anemone]

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12$.

Find $|f(0)|$.

 

[kaliprasad]

Spoiler: my Solution

Because a cubic polynomial shall have maximum 2 humps ( that is where it goes from down to up and from
up to down ) we shall have $f(1) = f(5)= f(6)$ and $f(2) = f(3) = f(7)$ and $f(1) \ne f(2)$.

For them to have values not meeting this criteria we shall have more humps .

As we are interested in absolute value we can choose $f(1) = - f(2) = 12$

As $f(1) = f(5)= f(6) = 12$ So we have $f(1) - 12 = f(5) - 12 = f(6) - 12 = 0$

So we can choose $f(x) = A(x-1)(x-5)(x-6) + 12 $ where A is a constant whose value we need to find

Putting $x=2$ we get $f(2) = A* 1 * (-3) *(-4) + 12 = 12A + 12 = - 12$ or A = -2

So $f(x) = -2(x-1)(x-5)(x-6) + 12$

Or $f(0) = -2 * (-1)(-5)(-6) + 12 = 72$

Or $\left | f(0) \right |= 72 $