mathcalculus
locate the critical points: g(x)= (1*x+1)/17x



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mathcalculus
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so far im left with : 7x^27x6

mathcalculus
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i know that i have to = to zero

mathcalculus
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then I'm stuck.

mathcalculus
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factor?

Peter14
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could you write out the equation more clearly?

mathcalculus
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g(x) \[\frac{ 1*x+1 }{17x}\]

Peter14
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then you use quotient rule?

mathcalculus
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yes

mathcalculus
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i have 7x^27x6

mattt9
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dw:1363323083958:dw

mathcalculus
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i found the derivative..

mattt9
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dw:1363323153025:dw

mathcalculus
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@mattt9 the derivative that i got is this: \[\frac{ 1*(17x)7(x+1) }{ (17x)^{2} }\]

mattt9
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dw:1363323222354:dw

mathcalculus
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how did i get the derivative wrong?

mattt9
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you missed a negative

Peter14
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matt, I don't think you can take away the ^2 like that...

mattt9
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hmm i think you're right actually.

mathcalculus
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the answer is none

mattt9
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okay that's what i thought.

mattt9
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after peter reclarified that for me

mathcalculus
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x cannot be 1/7 because you cancel the 7x+7x and youre left with 6

Peter14
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graphing\[y=\frac{ 1 }{ (17x)^2 }\] with Desmos I don't see a place where it crosses the Xaxis

mathcalculus
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when you multiply the denominator to zero= 0

mathcalculus
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and 6 does not equal to an x

Peter14
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thinking about it: 1/anything cannot equal 0

Peter14
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so there are no critical points