My Math Forum Question regarding a critical point formula

 Calculus Calculus Math Forum

 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 $ax^3 + bx^2 + cx + d$, and I won't be able to find the critical points. I know that when you have the form $ax^2 + bx + c$, you can use $(-b +/- sqrt(b^2-4ac))/(2a)$. Is there such a method for $ax^3 + bx^2 + cx + d$ ?
 August 20th, 2013, 04:58 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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(4b²-12ac)]/6a"
August 21st, 2013, 09:13 AM   #5
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,879
Thanks: 1087

Math Focus: Elementary mathematics and beyond
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 x³ + x² + 5x + 3, for example.

August 21st, 2013, 09:48 AM   #6
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: Question regarding a critical point formula

Quote:
 Originally Posted by John Marsh "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(4b²-12ac)]/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, 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 $x\,=\,\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}$. 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.

 Tags critical, formula, point, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post syamilsynz Applied Math 2 April 5th, 2013 06:36 PM felicia184 Calculus 5 September 17th, 2012 10:05 AM Mike86 Applied Math 2 October 9th, 2010 06:02 AM ManuLi Algebra 3 August 20th, 2009 02:25 AM felicia184 Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top