## anonymous one year ago Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When g(t) = t+ cost , -2pi less than or equal to "t" less than or equal to 2pi A) Find the intervals of increase of decrease B) Find the local max and mins C) Find the intervals of concavity and inflection points D) Use the information from parts A-C to sketch the graph.

1. anonymous

So for the first derivative I got: g' (t) = 1-sint t and so the critical values would be pi/2 & -pi/2 ?

2. DanJS

sin(x) = 1 when x = ... in the interval of the function

3. anonymous

when x= -pi/2 and pi/2 ?

4. DanJS

just pi/2 so far

5. DanJS

-pi/2 is the negative y axis and the sin would be -1

6. DanJS

so you can subtract 2 pi from pi/2 and still be in the interval of the function

7. anonymous

hmm... so its jut pi/2? I get really confused when they start given me intervals where there is negatives? Does -2pi mean we go around the unit circle clockwise?

8. DanJS

sin(x) = 1 , when x = pi/2 + k*2pi, for any k value

9. DanJS

just swinging around 360 degrees on the unit circle to the same spot with the 2pi

10. DanJS

when k=-1 x = p/2 - 2 pi = -3pi/2

11. DanJS

so you have two critical points inside the interval of the function.... x = pi/2 and x = -3pi/2

12. anonymous

So to test out a number before -3pi/2 can I choose -pi?

13. DanJS

no , larger, -3/2 is -1.5

14. DanJS

try -2pi

15. anonymous

That would give me a postive value when plugged in

16. DanJS

right, do the same for the other 2 intervals...

17. anonymous

Could I plug in 0 for the numbers between -3pi/2 and pi/2 and then 2pi after pi/2?

18. DanJS

yes of course

19. anonymous

sorry, just making sure

20. DanJS

any number inside the interval will work ,

21. anonymous

thats weird, all of them give me postive values? So does that mean that there is no interval where the function is decreasing?

22. DanJS

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23. DanJS

yep, the function is always increasing over the interval, but they do have horizontal tangents at those critical values like above

24. DanJS

calculate the y values, so you can add the (x,y) points for those two points on the graph

25. DanJS

since it is always increasing you can sketch a rough draft , you may want to calculate the x and y intercept points also

26. DanJS

|dw:1437609016552:dw|

27. DanJS

The points are like the horizontal tangent on the graph of y=x^3, increasing both sides

28. anonymous

I just dont get how you got the horizontal tangent?

29. DanJS

you took the derivative, and found that sin(x) = 1 , when the derivative is zero right?

30. anonymous

OH.. yea.. those values we got were the horizontal tangent ...

31. DanJS

yes, when you hear derivative, think "slope of a tangent line there"

32. anonymous

Yea, just forgot for a second there

33. anonymous

So is function increasing from (-2pi, 2pi)?

34. DanJS

no problem, so simple things to add to the graph to refine before the second derivative are, x and y intercepts, and the values of the function at the endpoints of the interval

35. DanJS

right, increasing everywhere

36. DanJS

with horizontal tangents at those two pooints you found

37. anonymous

so for the horizontal tangents I got points: (-pi/2, - pi/2) and the (-3pi/2, -5.712)

38. anonymous

and then for the second derivative, I would get: g''(t) = cos (t)

39. anonymous

And for its critical values I would get pi/2 and -3pi/2, the same values we got before for the first derivative?

40. DanJS

41. DanJS

Those are all the points you can add before the second derivative test

42. DanJS

any up to there you dont understand

43. anonymous

Okay, and well when did the second derivative test.. I got CD: (-2pi, -3pi/2) and CD: (-3pi/2, pi/2) U (pi/2,2pi)

44. anonymous

For the inflection point: I got (-3pi/2, -3pi/2)

45. anonymous

I have a good feeling im right

46. DanJS

$F(x) = x + \cos(x)$ $F'(x) = 1 + \sin(x)$ $F''(x) = -\cos(x)$

47. DanJS

second derivative is negative cosine

48. anonymous

wait...but doesnt -sin = cos?

49. anonymous

so that would just make it cos(t)

50. DanJS

sorry back

51. anonymous

lol k you almost gave me a heartattack

52. DanJS

yes, so cos(x) = 0 at pi/2 and 3pi/2 plus and minus since you can move 2 pi backwards

53. anonymous

we dont use -3pi/2? Only 3pi/2?

54. DanJS

55. DanJS

notice where cos(x) is zero in the interval

56. anonymous

wow, so do I use all those points?

57. DanJS

Cos(x) = 0 when the angle is at 90 or 270 , or any addition or subtraction of 360

58. DanJS

Cos(x) = 0 , x = pi/2 , x = 3pi/2, take off 2 pi from both x = -pi/2, x = -3pi/2

59. DanJS

It is the X coordinate on the unit circle, zero on the Y axis at 90 and 270 degrees

60. anonymous

yea makes sense

61. DanJS

So those are your 4 test points

62. anonymous

k thanks

63. DanJS

welcome, the graph of the actual function is above for reference..