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November 5th, 2013, 05:33 PM  #1 
Member Joined: Sep 2013 Posts: 58 Thanks: 0  Critical Points of a Rational Function
Okay, so today on my test under the Thinking section I was given this Eqn: F(x)=(12x^21)/x^3 It asked to find the critical points of the function I got Irocf(x)=24x/3x^2 Subbed Iroc f(x)=0 Solved x=0, plugged it into f(x) and got 1/0. So I wrote there is no Critical point (Point where slope of the tangent is 0) Im not sure if I found IROC f(x) right in the first place. Can anyone help? Thanks, Tzad 
November 5th, 2013, 05:36 PM  #2 
Member Joined: Sep 2013 Posts: 58 Thanks: 0  Re: Critical Points of a Rational Function
Can you find the Iroc of the Denominator, then the numerator seperately like I did above? When I treated it as 1, I got some HUUUUGGEEEE 12x^5...... thing all over hx(x+h)^3 (using first principles definition). Thanks, Tzad 
November 5th, 2013, 06:28 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: Critical Points of a Rational Function 
November 5th, 2013, 06:38 PM  #4  
Member Joined: Sep 2013 Posts: 58 Thanks: 0  Re: Critical Points of a Rational Function Quote:
 
November 5th, 2013, 06:43 PM  #5 
Member Joined: Sep 2013 Posts: 58 Thanks: 0  Re: Critical Points of a Rational Function
I still don't know how to get that though. Im kind of mad though, I asked my teacher is there a way to find the coordinates of those bumps, and he said "At your level no." Then he puts it on the test. From, Tzad 
November 6th, 2013, 04:35 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: Critical Points of a Rational Function
We have Let's find the derivative of this using first principles. Generally, 
November 6th, 2013, 01:14 PM  #7 
Member Joined: Sep 2013 Posts: 58 Thanks: 0  Re: Critical Points of a Rational Function
Thanks Greg, this question mucked me on the test. I screwed up First Principles .' Now I know how to do it. So this is the short way of doing it, I didn't make the connection that I need to do x^3, instead I did both the numerator and denominator separately like an idiot =(12x^21)/(x^3) =x^3(12x^21) =12x^1x^3 =12x^2+3x^4 =x^4(12x^2+3) Irocf(x)=(12x^2+3)/x^4 Btw, I highly recommend a donation option for this forum! I'd be glad to donate to keep it running! 
November 7th, 2013, 02:03 PM  #8 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Critical Points of a Rational Function
I would like to know what "IROC" means! Whatever it means, "critical points" of function, f(x), are points at which f'(x)= 0 or f'(x) does not exist. And the derivative of a quotient, , is [latex]\dfrac{f'(x)g(x) f(x)g'(x)}{g^2(x)} NOT . 
November 7th, 2013, 07:55 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: Critical Points of a Rational Function
Instantaneous rate of change.


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