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March 2, 2015

March 2, 2015

Posted by **Riley** on Tuesday, April 7, 2009 at 10:01am.

I have no idea how to solve this. Could you please give step by step answers and explanations for the steps. Thanks!

- calculus -
**Count Iblis**, Tuesday, April 7, 2009 at 10:15amThe local extrema correspond to zeroes of the derivative. That derivative is a wauadratic function. So, you should write down a quadratic function that has its zeroes at x = -2 and x = 1. Then integrate that function. This function is fixed up to an overall factor.

Then you integrate this function and fix the overall factor and the integration constant by demanding that the correct values are assumed at the local extrema.

- calculus-answer?? -
**Riley**, Tuesday, April 7, 2009 at 10:56amso

f(1)=a+b+c+d=0

f(-2)=-8a+4b-2c+d=3

f'(1)=3a+2b+c=0

f'(-2)=12a-4b+c=0

Now do I solve the system of equations for a,b,c,d? But is it possible to solve for this system of equations with 4 unknowns? Can I only substitute derivative into derivative and original into original or does it not matter?

- calculus -
**bobpursley**, Tuesday, April 7, 2009 at 10:58amyou can solve them anyway you want. I would use matrix algebra.

- calculus -
**Count Iblis**, Tuesday, April 7, 2009 at 11:09amI would suggest you start with this:

f'(x) = C(x+2)(x-1)

Then what is the general form of f(x)?

- calculus-how to solve? -
**Riley**, Tuesday, April 7, 2009 at 5:26pmCould you explain how to solve? How would you do matrix algebra here?

- calculus -
**Count Iblis**, Tuesday, April 7, 2009 at 5:40pmf'(x) = C(x+2)(x-1) =

C(x^2 + x - 2)

f(x) = C(1/3 x^3 + 1/2 x^2 - 2 x) + p

We need to satisfy the equations:

f(-2) = 3

f(1) = 0

So, we have:

10/3 C + p = 3

-7/6 C + p = 0

The solution is:

C = 2/3 and p = 7/9

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