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August 20th, 2013, 04:26 AM   #1
MFP
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Question regarding a critical point formula

Hey everyone,

I fear that on my final I'll be given an equation that ressembles , and I won't be able to find the critical points.

I know that when you have the form , you can use .

Is there such a method for ?
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August 20th, 2013, 04:58 AM   #2
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Re: Question regarding a critical point formula

There is a cubic formula

http://en.m.wikipedia.org/wiki/Cubic_function

But it is a bit complicated to use so even if you get a cubic function it is better to use the techniques of precalculus to determine rational roots (if any).

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August 20th, 2013, 08:47 AM   #3
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Re: Question regarding a critical point formula

ouch, I don't think I'll be able to memorize that formula in time haha. Alright, I'll use other methods. Thanks agentredlum!
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August 21st, 2013, 04:19 AM   #4
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Re: Question regarding a critical point formula

"Solution: Given ax+bx+cx+d
let f(x)= ax+bx+cx+d
f'(x)=3ax+2bx+c,where f'(x)=0
then there are two critical point ,and at least two real roots
there for we can use the formula x=[-2bv(4b-12ac)]/6a"
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August 21st, 2013, 09:13 AM   #5
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Re: Question regarding a critical point formula

Quote:
Originally Posted by John Marsh
then there are two critical point ,and at least two real roots
Not true. Consider x + x + 5x + 3, for example.
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August 21st, 2013, 09:48 AM   #6
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Re: Question regarding a critical point formula

Quote:
Originally Posted by John Marsh
"Solution: Given ax+bx+cx+d
let f(x)= ax+bx+cx+d
f'(x)=3ax+2bx+c,where f'(x)=0
then there are two critical point ,and at least two real roots
there for we can use the formula x=[-2bv(4b-12ac)]/6a"
I don't believe that was the question- although I thought it was at first, myself.

IF the original function were a cubic, then the derivative would be a quadratic but the OP ways he/she solved a quadratic before.
I would interpret this as asking what to do if the derivative itself is a cubic polynomial.

MFP, I doubt that you would be expected to know the cubic formula. IF the derivative turns out to be cubic, try simple integer values for x- bet on it being easily factorable.
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August 21st, 2013, 02:42 PM   #7
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Let c = p - q. Assume a ? 0, b, p and q are real numbers.
However if we have ax + bx + p = q then we have ax + bx + p - q = 0.
The typical quadratic formula would be .
To prove the quadratic formula is quite easy e.g. divide both sides by a then . . .
The quadratic formula is true if and only if ax + bx + c = 0.
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