## anonymous 4 years ago solve the following for x...

1. anonymous

$\frac{ f(x) - (-33) }{ x - 0 }$

2. anonymous

f(x) = $f(x) = 36xsin(17x) + 306\sqrt{3}x^2 -33$

3. anonymous

x=0

4. anonymous

$\frac{ f(x) - (-33)}{ x-0 } = f'(x)$

5. anonymous

6. anonymous

is it? the answers says that x = 5Pi/102 and 7Pi/102

7. anonymous

alright Plug in either 5pi/102 or 7pi/102 in f(x) and see if you get -33.

8. anonymous

the original question was find the x coordinates of all points on the curve $y = 36xsin(17x) + 306\sqrt{3}x^2 -33$ with a domain of $\frac{ \Pi }{ 34 } < x < \frac{ 3\Pi }{ 34 }$ where to tangent line passes through the point P(0, -33)

9. anonymous

so im not sure if the rise/run formula is the right thing to do

10. anonymous

also the point (0, -33) is actually not on the curve, but this line is touching points of the curve somewhere

11. anonymous

which is where answer x= 5Pi/102 and 7Pi/102 comes from

12. anonymous

alright. let's see if we can write the equation of the tangent line

13. anonymous

maybe?

14. anonymous

with one m being rise/ivrun and the other m being the derivativer equate the two toeeach other

15. anonymous

dont think it works though

16. anonymous

yea the derivative is not defined at 0

17. anonymous

m = m $\frac{ f(x) - (-33) }{ x - 0} = f'(x)$ $\frac{ (36xsin(17x) + 306\sqrt{3}x^2 -33) - (-33) }{x } = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x$

18. anonymous

i tried doing this but im still not getting the correct answer

19. anonymous

did you try getting f'(x), maybe i did it wrong

20. anonymous

is the x^2 in the square root?

21. anonymous

no

22. anonymous

oh, then I was doing it all wrong... let me start over

23. anonymous

the slope is 0; you just have to use the x in the denominator to cancel the xs in the numerator and take the limit of that as x approaches 0

24. anonymous

so take the limit after the rise/run formula after you factor it?

25. anonymous

of the rise/run*

26. anonymous

yea

27. anonymous

this gives you the slope, and then what do you do to get the answer

28. anonymous

$\lim_{x \rightarrow0}36\sin(17x) + 306\sqrt{3}x^2$

29. anonymous

slope equals to 0, and then...?

30. anonymous

0 = f'(x) 0 = $0 = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x$

31. anonymous

maybe solving for x here would be the answer?

32. anonymous

33. anonymous

oh did you get it?

34. anonymous

and 7pi/102... you took the derivative wrong

35. anonymous

oh..

36. anonymous

you shouldn't have 712, its 612

37. anonymous

what is f'(x)?

38. anonymous

ohh

39. anonymous

and set them equal as you did before, 36sin(17x)+306(3)^(1/2)x=36sin(17x)+612xcos(17x)+612sqrt(3)x

40. anonymous

solve for x, and thats what you get?

41. anonymous

$36\sin(17x)+306sqrt(3)x=36\sin(17x)+612xcos(17x)+612\sqrt(3)x$

42. anonymous

yea, x= (1/102)(12*k*pi+5pi) or x = (1/102)(12*k*pi-5pi)

43. anonymous

n is an integer, pick n so that x is between pi/34 and 3pi/34

44. anonymous

oh ok, thanks!

45. anonymous

np

46. anonymous

whats n?

47. anonymous

hold on a second I will attach a file soon

48. anonymous

there you go

49. anonymous

on the pdf file, it should not be 5pi + or - 12kpi; it should be 5pi+12kpi or x=-5pi+12kpi since cos(-a)=cos(a); it is 2kpi + or - a with a being 5pi/6 in our case... any questions??

50. anonymous

what is k?

51. anonymous

I used k instead of n; its the same thing

52. anonymous

oh ok

53. anonymous

when you isolated for k , what are you finding

54. anonymous

is that like the period

55. anonymous

no; im trying to find k. x should be in a certain range. so if x is between pi/34 and 3pi/34 then (pi/34)<x<(3pi/34). this will help you find an interval for k so that your x is in the right range or domain if you want to call it that... now k should be an integer so you look at the interval of k and decide which integers are inside that interval... if you do this for 12kpi-5pi; you should get k=1... pi/34<=(1/102)(12kpi-5pi)<=3pi/4... try it

56. anonymous

ok i finally got it

57. anonymous