## mathcalculus Group Title help: find the critical points: g(x) x+8/x-1 one year ago one year ago

1. mathcalculus Group Title

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

2. mathcalculus Group Title

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

3. mathcalculus Group Title

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

4. mathcalculus Group Title

then I'm stuck here :/

5. Twis7ed Group Title

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

6. Twis7ed Group Title

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 Group Title

yes so x=0?

8. mathcalculus Group Title

that's what I did later..

9. Twis7ed Group Title

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 Group Title

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

11. mathcalculus Group Title

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

12. mathcalculus Group Title

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

13. Twis7ed Group Title

What do you mean?

14. mathcalculus Group Title

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

15. mathcalculus Group Title

then there would be an x.

16. mathcalculus Group Title

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

17. Twis7ed Group Title

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 Group Title

whats the asymptote?

19. Twis7ed Group Title

At x=1

20. mathcalculus Group Title

how did you find 1 again?

21. mathcalculus Group Title

btw thank you

22. Twis7ed Group Title

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