Extra credits

Problem:



Solution:

$ f'(x) = 3ax^2 + 2bx + c $
$ f''(x) = 6ax + 2b $

The point of inflection has x coordinate -1, therefore

$ 6a(-1) + 2b = 0 $

The extremas has x coordinate 1 and -3, therefore, 1 and -3 are the roots of  $3ax^2 + 2bx + c = 0 $

$ 3a + 2b + c = 0 $
$ 3a(-3)^2 + 2b(-3) + c = 0 $.

The three equations tell us $ b = 3a, c = -9a $.

So the equation is $ f(x) = ax^3 + 3ax^2 - 9ax + d$

When x = 1, $ f(x) = a + 3a - 9a + d = -5a + d = -9 $
When x = -1, $ f(x) = -a + 3a + 9a + d = 11a + d = 23 $

Therefore $ a = 2, d = 1 $

The final answer is $ 2x^3 + 6x^2 - 18x + 1 $