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August 20th, 2013, 04:26 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, 04:58 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,276 Thanks: 204  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, 08:47 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, 04:19 AM  #4 
Newbie Joined: May 2013 Posts: 16 Thanks: 0  Re: Question regarding a critical point formula
"Solution: Given ax³+bx²+cx+d let f(x)= ax³+bx²+cx+d f'(x)=3ax²+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(4b²12ac)]/6a" 
August 21st, 2013, 09:13 AM  #5  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,516 Thanks: 910 Math Focus: Elementary mathematics and beyond  Re: Question regarding a critical point formula Quote:
 
August 21st, 2013, 09:48 AM  #6  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 5  Re: Question regarding a critical point formula Quote:
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, 02:42 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 ax² + bx + p = q then we have ax² + 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 ax² + bx + c = 0. 

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