## mathcalculus Group Title help! :( finding the critical point of g(x)= (x+4)/(x-1) 10 months ago 10 months ago

1. mathcalculus Group Title

i used the quotient rule!

2. mathcalculus Group Title

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

3. mathcalculus Group Title

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

4. satellite73 Group Title

i doubt there are any

5. mathcalculus Group Title

why?

6. satellite73 Group Title

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

7. satellite73 Group Title

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

8. mathcalculus Group Title

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

9. mathcalculus Group Title

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

10. satellite73 Group Title

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 Group Title

right

12. satellite73 Group Title

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

13. mathcalculus Group Title

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

14. mathcalculus Group Title

i know 1 gives you undefined.

15. satellite73 Group Title

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 Group Title

you mean the denominator?

17. satellite73 Group Title

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 Group Title

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

19. mathcalculus Group Title

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

20. satellite73 Group Title

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

21. mathcalculus Group Title

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 Group Title

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

23. satellite73 Group Title

when is a fraction equal to zero?

24. mathcalculus Group Title

oh no it doesn't.

25. mathcalculus Group Title

and 0 does not multiply -5..

26. satellite73 Group Title

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

27. mathcalculus Group Title

nope.

28. mathcalculus Group Title

i think i got it...

29. satellite73 Group Title

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

30. mathcalculus Group Title

thanks @satellite73 :)

31. satellite73 Group Title

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

32. mathcalculus Group Title

no

33. mathcalculus Group Title

1?

34. satellite73 Group Title

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

35. satellite73 Group Title

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

36. mathcalculus Group Title

right.

37. mathcalculus Group Title

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