## iop360 3 years ago solve the following for x...

1. iop360

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

2. iop360

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

3. josiahh

x=0

4. iop360

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

5. josiahh

6. iop360

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

7. josiahh

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

8. iop360

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. iop360

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

10. iop360

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

11. iop360

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

12. josiahh

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

13. iop360

maybe?

14. iop360

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

15. iop360

dont think it works though

16. josiahh

yea the derivative is not defined at 0

17. iop360

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. iop360

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

19. iop360

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

20. josiahh

is the x^2 in the square root?

21. iop360

no

22. josiahh

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

23. josiahh

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. iop360

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

25. iop360

of the rise/run*

26. josiahh

yea

27. iop360

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

28. iop360

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

29. iop360

slope equals to 0, and then...?

30. iop360

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

31. iop360

maybe solving for x here would be the answer?

32. josiahh

33. iop360

oh did you get it?

34. josiahh

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

35. iop360

oh..

36. josiahh

you shouldn't have 712, its 612

37. iop360

what is f'(x)?

38. iop360

ohh

39. josiahh

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

40. iop360

solve for x, and thats what you get?

41. josiahh

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

42. josiahh

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

43. josiahh

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

44. iop360

oh ok, thanks!

45. josiahh

np

46. iop360

whats n?

47. josiahh

hold on a second I will attach a file soon

48. josiahh

there you go

49. josiahh

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. iop360

what is k?

51. josiahh

I used k instead of n; its the same thing

52. iop360

oh ok

53. iop360

when you isolated for k , what are you finding

54. iop360

is that like the period

55. josiahh

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. iop360

ok i finally got it

57. iop360