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locate the critical points: g(x)= (1*x+1)/1-7x

Mathematics
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so far im left with : 7x^2-7x-6
i know that i have to = to zero
then I'm stuck.

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Other answers:

factor?
could you write out the equation more clearly?
g(x) \[\frac{ 1*x+1 }{1-7x}\]
then you use quotient rule?
yes
i have 7x^2-7x-6
|dw:1363323083958:dw|
i found the derivative..
|dw:1363323153025:dw|
@mattt9 the derivative that i got is this: \[\frac{ 1*(1-7x)-7(x+1) }{ (1-7x)^{2} }\]
|dw:1363323222354:dw|
how did i get the derivative wrong?
you missed a negative
matt, I don't think you can take away the ^-2 like that...
hmm i think you're right actually.
the answer is none
okay that's what i thought.
after peter reclarified that for me
x cannot be 1/7 because you cancel the -7x+7x and youre left with -6
graphing\[y=\frac{ 1 }{ (1-7x)^2 }\] with Desmos I don't see a place where it crosses the X-axis
when you multiply the denominator to zero= 0
and -6 does not equal to an x
thinking about it: 1/anything cannot equal 0
so there are no critical points

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