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November 5th, 2013, 05:33 PM   #1
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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?

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Tzad
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November 5th, 2013, 05:36 PM   #2
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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).

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Tzad
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November 5th, 2013, 06:28 PM   #3
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Re: Critical Points of a Rational Function

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November 5th, 2013, 06:38 PM   #4
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Re: Critical Points of a Rational Function

Quote:
Originally Posted by greg1313
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?
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November 5th, 2013, 06:43 PM   #5
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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.

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November 6th, 2013, 04:35 AM   #6
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Re: Critical Points of a Rational Function

We have

Let's find the derivative of this using first principles.



















Generally,









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November 6th, 2013, 01:14 PM   #7
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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!
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November 7th, 2013, 02:03 PM   #8
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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 .
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November 7th, 2013, 07:55 PM   #9
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Re: Critical Points of a Rational Function

Instantaneous rate of change.
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