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## anonymous 3 years ago help: find the critical points: g(x) x+8/x-1

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1. anonymous

$\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }$

2. anonymous

$\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }$

3. anonymous

$\frac{ -9 }{ x ^{2} -2x+1 }$

4. anonymous

then I'm stuck here :/

5. anonymous

Find x in: $\frac{ -9 }{ x^2-2x+1 } = 0$

6. 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.

7. anonymous

yes so x=0?

8. anonymous

that's what I did later..

9. 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.

10. anonymous

Also, since there is no solution for that equation (just realised :p) You can find at which point it is undefined.

11. anonymous

yes, i end up with -9=0 can you show me?

12. anonymous

it's undefined right? because the x is not there..

13. anonymous

What do you mean?

14. anonymous

unless it was something like -9x/ x^2-2x+1

15. anonymous

then there would be an x.

16. anonymous

but since -9 is alone... there is no critical point. right?

17. 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?

18. anonymous

whats the asymptote?

19. anonymous

At x=1

20. anonymous

how did you find 1 again?

21. anonymous

btw thank you

22. 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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