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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^2-1)/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 $\frac{d}{dx}\,\frac{12x^2\,-\,1}{x^3}\,=\,\frac{24x(x^3)\,-\,3x^2(12x^2\,-\,1)}{x^6}\,=\,\frac{24x^4\,-\,36x^4\,+\,3x^2}{x^6}\,=\,\frac{3\,-\,12x^2}{x^4}$
November 5th, 2013, 06:38 PM   #4
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Re: Critical Points of a Rational Function

Quote:
 Originally Posted by greg1313 $\frac{d}{dx}\,\frac{12x^2\,-\,1}{x^3}\,=\,\frac{24x(x^3)\,-\,3x^2(12x^2\,-\,1)}{x^6}\,=\,\frac{24x^4\,-\,36x^4\,+\,3x^2}{x^6}\,=\,\frac{3\,-\,12x^2}{x^4}$
I don't know how you got to this. Irocf(x)=Limh->0 [f(x+h)-f(x)]/h. Can you show it that way please?

 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 co-ordinates 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 $f(x)\,=\,\frac{12x^2\,-\,1}{x^3}$ Let's find the derivative of this using first principles. $\lim_{h\to0}\;\frac{\frac{12(x\,+\,h)^2\,-\,1}{(x\,+\,h)^3}\,-\,\frac{12x^2\,-\,1}{x^3}}{h}$ $=\,\lim_{h\to0}\;\frac{\frac{(12(x\,+\,h)^2\,-\,1)x^3}{x^3(x\,+\,h)^3}\,-\,\frac{(12x^2\,-\,1)(x\,+\,h)^3}{x^3(x\,+\,h)^3}}{h}$ $=\,\lim_{h\to0}\;\frac{(12x^2\,+\,24xh\,+\,12h^2\,-\,1)(x^3)\,-\,(12x^2\,-\,1)(x^3\,+\,3x^2h\,+\,3xh^2\,+\,h^3)}{h(x^3)(x\,+ \,h)^3}$ $=\,\lim_{h\to0}\;\frac{12x^5\,+\,24x^4h\,+\,12x^3h ^2\,-\,x^3\,-\,12x^5\,-\,36x^4h\,-\,36x^3h^2\,-\,12x^2h^3\,+\,x^3\,+\,3x^2h\,+\,3xh^2\,+\,h^3}{h( x^3)(x\,+\,h)^3}$ $=\,\lim_{h\to0}\;\frac{-12x^4h\,-\,24x^3h^2\,-\,12x^2h^3\,+\,3x^2h\,+\,3xh^2\,+\,h^3}{h(x^3)(x\, +\,h)^3}$ $=\,\lim_{h\to0}\;\frac{h(-12x^4\,-\,24x^3h\,-\,12x^2h^2\,+\,3x^2\,+\,3xh\,+\,h^2)}{h(x^3)(x\,+\ ,h)^3}$ $=\,\lim_{h\to0}\;\frac{-12x^4\,-\,24x^3h\,-\,12x^2h^2\,+\,3x^2\,+\,3xh\,+\,h^2}{(x^3)(x\,+\,h )^3}$ $=\,\frac{-12x^4\,+\,3x^2}{x^6}$ $=\,\frac{3\,-\,12x^2}{x^4}$ Generally, $\lim_{h\to0}\;\frac{\frac{f(x\,+\,h)}{g(x\,+\,h)}\ ,-\,\frac{f(x)}{g(x)}}{h}$ $=\,\lim_{x\to0}\;\frac{f(x\,+\,h)g(x)\,-\,f(x)g(x\,+\,h)}{hg(x)g(x\,+\,h)}$ $=\,\lim_{x\to0}\;\frac{f(x\,+\,h)g(x)\,-\,f(x)g(x\,+\,h)\,+\,f(x\,+\,h)g(x\,+\,h)\,-\,f(x\,+\,h)g(x\,+\,h)}{hg(x)g(x\,+\,h)}$ $=\,\lim_{x\to0}\;\frac{(f(x\,+\,h)\,-\,f(x))g(x)\,-\,f(x\,+\,h)(g(x\,+\,h)\,-\,g(x))}{hg(x)g(x\,+\,h)}$ $=\,\frac{f'(x)g(x)\,-\,f(x)g'(x)}{(g(x))^2}$
 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^2-1)/(x^3) =x^-3(12x^2-1) =12x^-1-x^-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, $\dfrac{f(x)}{g(x)}$, is [latex]\dfrac{f'(x)g(x)- f(x)g'(x)}{g^2(x)} NOT $\dfrac{f'(x)}{g'(x)}$.
 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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# finding the critical points on a rational function

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