## mathcalculus Group Title help! :( finding the critical point of g(x)= (x+4)/(x-1) 11 months ago 11 months ago

1. mathcalculus

i used the quotient rule!

2. mathcalculus

i got my critical point 5 and 1 but it's wrong :(

3. mathcalculus

-5/(x^2-2x+1)= 0

4. satellite73

i doubt there are any

5. mathcalculus

why?

6. satellite73

the derivative is $$\frac{-5}{(x-1)^2}$$

7. satellite73

a fraction is only zero if the numerator is zero this numerator is $$-5$$

8. mathcalculus

-5/(x^2-2x+1)= 0

9. mathcalculus

right so when i get to this: -5/(x^2-2x+1)= 0 what do i do?

10. satellite73

the denominator is positive for all values of $$x$$ except for $$x=1$$ where both it, and the original function are undefined the numerator is negative that tells you the function is always decreasing

11. mathcalculus

right

12. satellite73

you may remember what such a thing looks like from a pre - calc course

13. mathcalculus

ok but then why isn't -5 considered a critical point?

14. mathcalculus

i know 1 gives you undefined.

15. satellite73

a critical point is not the numerator a critical point is a number $$a$$ where either $$f'(a)=0$$ or $$f'(a)$$ does not exist

16. mathcalculus

you mean the denominator?

17. satellite73

graph your derivative you will see that it never crosses the $$y$$ axis, that it always lives below it i.e. it is always negative if you graph your original function, you will see that it is always decreasing

18. mathcalculus

i did, but what i don't understand why it has no critical point.

19. mathcalculus

okay but in this question im not allowed to use the calculator.

20. satellite73

well, because it doesn't does $$f(x)=2x+1$$ have a critical point?

21. mathcalculus

so if i can't graph it, how do can i automatically tell there is no critical point by looking at the derivative -5/(x^2-2x+1)= 0

22. satellite73

of course it doesn't, because for $$f(x)=2x+1$$you have a line that is always increasing

23. satellite73

when is a fraction equal to zero?

24. mathcalculus

oh no it doesn't.

25. mathcalculus

and 0 does not multiply -5..

26. satellite73

is $$\frac{5}{x}=0$$ solvable?

27. mathcalculus

nope.

28. mathcalculus

i think i got it...

29. satellite73

right, and neither is $$\frac{-5}{(x-1)^2}$$

30. mathcalculus

thanks @satellite73 :)

31. satellite73

how about $$e^x$$? is that ever zero? yw

32. mathcalculus

no

33. mathcalculus

1?

34. satellite73

no, $$e^x>0$$ for all $$x$$ $$e^1=e$$ not zero

35. satellite73

the function $$f(x)=e^x$$ is strictly increasing.

36. mathcalculus

right.

37. mathcalculus

@satellite73 i have a question: how come -5/(x-1)=0 is not a critical point and -4/15 is?