A community for students.
Here's the question you clicked on:
 0 viewing
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 AC to sketch the graph.
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 AC to sketch the graph.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for the first derivative I got: g' (t) = 1sint t and so the critical values would be pi/2 & pi/2 ?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0sin(x) = 1 when x = ... in the interval of the function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when x= pi/2 and pi/2 ?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0pi/2 is the negative y axis and the sin would be 1

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0so you can subtract 2 pi from pi/2 and still be in the interval of the function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm... 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?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0sin(x) = 1 , when x = pi/2 + k*2pi, for any k value

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0just swinging around 360 degrees on the unit circle to the same spot with the 2pi

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0when k=1 x = p/2  2 pi = 3pi/2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0so you have two critical points inside the interval of the function.... x = pi/2 and x = 3pi/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So to test out a number before 3pi/2 can I choose pi?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0no , larger, 3/2 is 1.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would give me a postive value when plugged in

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0right, do the same for the other 2 intervals...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could I plug in 0 for the numbers between 3pi/2 and pi/2 and then 2pi after pi/2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, just making sure

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0any number inside the interval will work ,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats weird, all of them give me postive values? So does that mean that there is no interval where the function is decreasing?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yep, the function is always increasing over the interval, but they do have horizontal tangents at those critical values like above

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0calculate the y values, so you can add the (x,y) points for those two points on the graph

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0since it is always increasing you can sketch a rough draft , you may want to calculate the x and y intercept points also

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0The points are like the horizontal tangent on the graph of y=x^3, increasing both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just dont get how you got the horizontal tangent?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0you took the derivative, and found that sin(x) = 1 , when the derivative is zero right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH.. yea.. those values we got were the horizontal tangent ...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, when you hear derivative, think "slope of a tangent line there"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, just forgot for a second there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is function increasing from (2pi, 2pi)?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0no 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

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0right, increasing everywhere

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0with horizontal tangents at those two pooints you found

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for the horizontal tangents I got points: (pi/2,  pi/2) and the (3pi/2, 5.712)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then for the second derivative, I would get: g''(t) = cos (t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And for its critical values I would get pi/2 and 3pi/2, the same values we got before for the first derivative?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Those are all the points you can add before the second derivative test

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0any up to there you dont understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, and well when did the second derivative test.. I got CD: (2pi, 3pi/2) and CD: (3pi/2, pi/2) U (pi/2,2pi)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the inflection point: I got (3pi/2, 3pi/2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have a good feeling im right

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[F(x) = x + \cos(x)\] \[F'(x) = 1 + \sin(x)\] \[F''(x) = \cos(x) \]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0second derivative is negative cosine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait...but doesnt sin = cos?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that would just make it cos(t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol k you almost gave me a heartattack

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, so cos(x) = 0 at pi/2 and 3pi/2 plus and minus since you can move 2 pi backwards

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we dont use 3pi/2? Only 3pi/2?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0notice where cos(x) is zero in the interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow, so do I use all those points?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Cos(x) = 0 when the angle is at 90 or 270 , or any addition or subtraction of 360

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Cos(x) = 0 , x = pi/2 , x = 3pi/2, take off 2 pi from both x = pi/2, x = 3pi/2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0It is the X coordinate on the unit circle, zero on the Y axis at 90 and 270 degrees

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0So those are your 4 test points

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0welcome, the graph of the actual function is above for reference..
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.