## anonymous 3 years ago locate the critical points: g(x)= (1*x+1)/1-7x

1. anonymous

so far im left with : 7x^2-7x-6

2. anonymous

i know that i have to = to zero

3. anonymous

then I'm stuck.

4. anonymous

factor?

5. anonymous

could you write out the equation more clearly?

6. anonymous

g(x) $\frac{ 1*x+1 }{1-7x}$

7. anonymous

then you use quotient rule?

8. anonymous

yes

9. anonymous

i have 7x^2-7x-6

10. anonymous

|dw:1363323083958:dw|

11. anonymous

i found the derivative..

12. anonymous

|dw:1363323153025:dw|

13. anonymous

@mattt9 the derivative that i got is this: $\frac{ 1*(1-7x)-7(x+1) }{ (1-7x)^{2} }$

14. anonymous

|dw:1363323222354:dw|

15. anonymous

how did i get the derivative wrong?

16. anonymous

you missed a negative

17. anonymous

matt, I don't think you can take away the ^-2 like that...

18. anonymous

hmm i think you're right actually.

19. anonymous

20. anonymous

okay that's what i thought.

21. anonymous

after peter reclarified that for me

22. anonymous

x cannot be 1/7 because you cancel the -7x+7x and youre left with -6

23. anonymous

graphing$y=\frac{ 1 }{ (1-7x)^2 }$ with Desmos I don't see a place where it crosses the X-axis

24. anonymous

when you multiply the denominator to zero= 0

25. anonymous

and -6 does not equal to an x

26. anonymous

thinking about it: 1/anything cannot equal 0

27. anonymous

so there are no critical points