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April 18, 2014

April 18, 2014

Posted by **kooldude** on Thursday, November 9, 2006 at 10:18am.

First you calculate the derivative:

f'(x)=6x^2 + 2ax + b (1)

At the local maximum and minumum f' is zero. If a polynomial is zero at some point p, then it must contain a factor (x-p). So, since you know that f' is zero at both x =-4 and x = 5 you know that f' must be of the form:

f'(x) = A(x+4)(x-5) (2)

the two factors (x+4) and (x-5) make the right hand side a second degree polynomial, so A must be a constant. If f' were a third degree function then A would have been an unknown linea function. From (1) and (2) you find

6x^2 + 2ax + b = A (x+4)(x-5)

The coefficient of x^2 on the left hand side is 6, on the right hand side it is A, so you find that A = 6.

6 (x+4)(x-5) = 6 x^2 -6 x -120,

so 2a = -6 --> a = -3

and b = -120.

thanks!

it would help if you wrote the quistionbozo and by the way YOUR NOT KOOL

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