## mathcalculus 2 years ago help: find the critical points: g(x) x+8/x-1

1. mathcalculus

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

2. mathcalculus

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

3. mathcalculus

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

4. mathcalculus

then I'm stuck here :/

5. Twis7ed

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

6. Twis7ed

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. mathcalculus

yes so x=0?

8. mathcalculus

that's what I did later..

9. Twis7ed

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. Twis7ed

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

11. mathcalculus

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

12. mathcalculus

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

13. Twis7ed

What do you mean?

14. mathcalculus

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

15. mathcalculus

then there would be an x.

16. mathcalculus

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

17. Twis7ed

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. mathcalculus

whats the asymptote?

19. Twis7ed

At x=1

20. mathcalculus

how did you find 1 again?

21. mathcalculus

btw thank you

22. Twis7ed

Set the bottom of the function equal to zero so x^2-2x+1 = 0 and solve for x. No problem :D