Expert: Scotto - 5/14/2009

Question
how do you solve this?

find a cubic functions that had a local maximum value of 3 at -2 and a local minimum value of 0 at 1.

Answer
This means that the derivative is 0 at x = -2 and x = 1.
hat will say that the derivative is of the form M(x+2)(x-1).
Multiplying it out gives M(x²+x-2).  Integrating to get the function gives M(x^3/3 + x²/2 - 2x) + K where M and K are both constants.

What we know is the value of the function at each of the two points.
At x=-2, the value is 3.  This says that M(-8/3 + 4/2 + 4) + K = 3.
At x=1, the value is 0.  This says that M(1/3 + 1/2 - 2) + K = 0.

We can reduce the expression in the first equation.
It can be seen that (-8/3 + 4/2 + 4) is really
(-16/6 + 12/6 + 24/6) =20/6.  
The equation is then 20M/6 + K = 3.

We can reduce the second equation by noting that
(1/3 + 1/2 - 2) = 2/6 + 3/6 - 12/6 = -7/6.
This equation is now -7M/6 + K = 0.

The two eqations can both easily be solved for K and then they can be set equal to each other.

That would gives 3 - 20M/6 = 7M/6 => 3 = 27M/6.
That means that M is 6*3/27, or 18/27, or 2/3.
This would say that K = (7/6)(2/3) = 7/9.

Giving the values of M and K, the equation would be
f(x) = (2/3)(x^3/3 + x²/2 - 2x) + 7/9.

Multiplying out the equation gives
f(x) = 2x^3/9 + 2x²/6 - 4x/3 + 7/9.

A common denominator can be seen to be 9.  
2x^3/9 is OK since its over 9 already.
It can be seen that 2x²/6 = 3x²/9.
It can be seen that -4x/3 = -12x/9.
7/9 is already over 9.

We now know that f(x) = (2x^3 + 3x² - 12x + 7)/9.