## anonymous 2 years ago help! :( finding the critical point of g(x)= (x+4)/(x-1)

1. anonymous

i used the quotient rule!

2. anonymous

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

3. anonymous

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

4. anonymous

i doubt there are any

5. anonymous

why?

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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. anonymous

right

12. anonymous

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

13. anonymous

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

14. anonymous

i know 1 gives you undefined.

15. anonymous

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. anonymous

you mean the denominator?

17. anonymous

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. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

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. anonymous

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

23. anonymous

when is a fraction equal to zero?

24. anonymous

oh no it doesn't.

25. anonymous

and 0 does not multiply -5..

26. anonymous

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

27. anonymous

nope.

28. anonymous

i think i got it...

29. anonymous

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

30. anonymous

thanks @satellite73 :)

31. anonymous

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

32. anonymous

no

33. anonymous

1?

34. anonymous

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

35. anonymous

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

36. anonymous

right.

37. anonymous

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