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 August 20th, 2013, 05:26 AM #1 Newbie   Joined: Mar 2011 Posts: 17 Thanks: 0 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 ? August 20th, 2013, 05:58 AM #2 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 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).  August 20th, 2013, 09:47 AM #3 Newbie   Joined: Mar 2011 Posts: 17 Thanks: 0 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! August 21st, 2013, 05:19 AM #4 Newbie   Joined: May 2013 Posts: 16 Thanks: 0 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=[-2b�v(4b�-12ac)]/6a" August 21st, 2013, 10:13 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. August 21st, 2013, 10:48 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=[-2b�v(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. August 21st, 2013, 03:42 PM #7 Member   Joined: Aug 2013 Posts: 59 Thanks: 0 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. Tags critical, formula, point, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post syamilsynz Applied Math 2 April 5th, 2013 07:36 PM felicia184 Calculus 5 September 17th, 2012 11:05 AM Mike86 Applied Math 2 October 9th, 2010 07:02 AM ManuLi Algebra 3 August 20th, 2009 03:25 AM felicia184 Algebra 0 December 31st, 1969 04:00 PM

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