## mathcalculus Group Title locate the critical points: g(x)= (1*x+1)/1-7x one year ago one year ago

1. mathcalculus

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

2. mathcalculus

i know that i have to = to zero

3. mathcalculus

then I'm stuck.

4. mathcalculus

factor?

5. Peter14

could you write out the equation more clearly?

6. mathcalculus

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

7. Peter14

then you use quotient rule?

8. mathcalculus

yes

9. mathcalculus

i have 7x^2-7x-6

10. mattt9

|dw:1363323083958:dw|

11. mathcalculus

i found the derivative..

12. mattt9

|dw:1363323153025:dw|

13. mathcalculus

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

14. mattt9

|dw:1363323222354:dw|

15. mathcalculus

how did i get the derivative wrong?

16. mattt9

you missed a negative

17. Peter14

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

18. mattt9

hmm i think you're right actually.

19. mathcalculus

20. mattt9

okay that's what i thought.

21. mattt9

after peter reclarified that for me

22. mathcalculus

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

23. Peter14

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

24. mathcalculus

when you multiply the denominator to zero= 0

25. mathcalculus

and -6 does not equal to an x

26. Peter14

thinking about it: 1/anything cannot equal 0

27. Peter14

so there are no critical points