## iop360 Group Title solve the following for x... one year ago one year ago

1. iop360 Group Title

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

2. iop360 Group Title

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

3. josiahh Group Title

x=0

4. iop360 Group Title

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

5. josiahh Group Title

that is the derivative formula at x=0; X=0 is your answer

6. iop360 Group Title

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

7. josiahh Group Title

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

8. iop360 Group Title

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 Group Title

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

10. iop360 Group Title

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

11. iop360 Group Title

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

12. josiahh Group Title

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

13. iop360 Group Title

maybe?

14. iop360 Group Title

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

15. iop360 Group Title

dont think it works though

16. josiahh Group Title

yea the derivative is not defined at 0

17. iop360 Group Title

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 Group Title

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

19. iop360 Group Title

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

20. josiahh Group Title

is the x^2 in the square root?

21. iop360 Group Title

no

22. josiahh Group Title

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

23. josiahh Group Title

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 Group Title

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

25. iop360 Group Title

of the rise/run*

26. josiahh Group Title

yea

27. iop360 Group Title

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

28. iop360 Group Title

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

29. iop360 Group Title

slope equals to 0, and then...?

30. iop360 Group Title

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

31. iop360 Group Title

maybe solving for x here would be the answer?

32. josiahh Group Title

the answer is 5pi/102

33. iop360 Group Title

oh did you get it?

34. josiahh Group Title

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

35. iop360 Group Title

oh..

36. josiahh Group Title

you shouldn't have 712, its 612

37. iop360 Group Title

what is f'(x)?

38. iop360 Group Title

ohh

39. josiahh Group Title

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

40. iop360 Group Title

solve for x, and thats what you get?

41. josiahh Group Title

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

42. josiahh Group Title

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

43. josiahh Group Title

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

44. iop360 Group Title

oh ok, thanks!

45. josiahh Group Title

np

46. iop360 Group Title

whats n?

47. josiahh Group Title

hold on a second I will attach a file soon

48. josiahh Group Title

there you go

49. josiahh Group Title

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 Group Title

what is k?

51. josiahh Group Title

I used k instead of n; its the same thing

52. iop360 Group Title

oh ok

53. iop360 Group Title

when you isolated for k , what are you finding

54. iop360 Group Title

is that like the period

55. josiahh Group Title

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 Group Title

ok i finally got it

57. iop360 Group Title

thanks for your help!

58. josiahh Group Title

np...