help: find the critical points: g(x) x+8/x-1

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help: find the critical points: g(x) x+8/x-1

Calculus1
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\[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]
\[\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }\]
\[\frac{ -9 }{ x ^{2} -2x+1 }\]

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then I'm stuck here :/
Find x in: \[\frac{ -9 }{ x^2-2x+1 } = 0\]
Remember that the value of the derivative is the slope of the line of g(x) at that point so when the first derivative is equal to zero, the slope of the curve is zero and it is therefore a critical point.
yes so x=0?
that's what I did later..
No, when x = 0 then the function is equal to -9...find the value of x so that the dericative is equal to 0.
Also, since there is no solution for that equation (just realised :p) You can find at which point it is undefined.
yes, i end up with -9=0 can you show me?
it's undefined right? because the x is not there..
What do you mean?
unless it was something like -9x/ x^2-2x+1
then there would be an x.
but since -9 is alone... there is no critical point. right?
Correct, but there IS an asymptote in the graph so I'm not sure if you'd consider that a critical point in your class?
whats the asymptote?
At x=1
how did you find 1 again?
btw thank you
Set the bottom of the function equal to zero so x^2-2x+1 = 0 and solve for x. No problem :D

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