anonymous
  • anonymous
help: find the critical points: g(x) x+8/x-1
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]
anonymous
  • anonymous
\[\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }\]
anonymous
  • anonymous
\[\frac{ -9 }{ x ^{2} -2x+1 }\]

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anonymous
  • anonymous
then I'm stuck here :/
anonymous
  • anonymous
Find x in: \[\frac{ -9 }{ x^2-2x+1 } = 0\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes so x=0?
anonymous
  • anonymous
that's what I did later..
anonymous
  • anonymous
No, when x = 0 then the function is equal to -9...find the value of x so that the dericative is equal to 0.
anonymous
  • anonymous
Also, since there is no solution for that equation (just realised :p) You can find at which point it is undefined.
anonymous
  • anonymous
yes, i end up with -9=0 can you show me?
anonymous
  • anonymous
it's undefined right? because the x is not there..
anonymous
  • anonymous
What do you mean?
anonymous
  • anonymous
unless it was something like -9x/ x^2-2x+1
anonymous
  • anonymous
then there would be an x.
anonymous
  • anonymous
but since -9 is alone... there is no critical point. right?
anonymous
  • anonymous
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?
anonymous
  • anonymous
whats the asymptote?
anonymous
  • anonymous
At x=1
anonymous
  • anonymous
how did you find 1 again?
anonymous
  • anonymous
btw thank you
anonymous
  • anonymous
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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