anonymous
  • anonymous
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.
Calculus1
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
So for the first derivative I got: g' (t) = 1-sint t and so the critical values would be pi/2 & -pi/2 ?
DanJS
  • DanJS
sin(x) = 1 when x = ... in the interval of the function
anonymous
  • anonymous
when x= -pi/2 and pi/2 ?

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DanJS
  • DanJS
just pi/2 so far
DanJS
  • DanJS
-pi/2 is the negative y axis and the sin would be -1
DanJS
  • DanJS
so you can subtract 2 pi from pi/2 and still be in the interval of the function
anonymous
  • 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?
DanJS
  • DanJS
sin(x) = 1 , when x = pi/2 + k*2pi, for any k value
DanJS
  • DanJS
just swinging around 360 degrees on the unit circle to the same spot with the 2pi
DanJS
  • DanJS
when k=-1 x = p/2 - 2 pi = -3pi/2
DanJS
  • DanJS
so you have two critical points inside the interval of the function.... x = pi/2 and x = -3pi/2
anonymous
  • anonymous
So to test out a number before -3pi/2 can I choose -pi?
DanJS
  • DanJS
no , larger, -3/2 is -1.5
DanJS
  • DanJS
try -2pi
anonymous
  • anonymous
That would give me a postive value when plugged in
DanJS
  • DanJS
right, do the same for the other 2 intervals...
anonymous
  • anonymous
Could I plug in 0 for the numbers between -3pi/2 and pi/2 and then 2pi after pi/2?
DanJS
  • DanJS
yes of course
anonymous
  • anonymous
sorry, just making sure
DanJS
  • DanJS
any number inside the interval will work ,
anonymous
  • anonymous
thats weird, all of them give me postive values? So does that mean that there is no interval where the function is decreasing?
DanJS
  • DanJS
|dw:1437608804982:dw|
DanJS
  • DanJS
yep, the function is always increasing over the interval, but they do have horizontal tangents at those critical values like above
DanJS
  • DanJS
calculate the y values, so you can add the (x,y) points for those two points on the graph
DanJS
  • DanJS
since it is always increasing you can sketch a rough draft , you may want to calculate the x and y intercept points also
DanJS
  • DanJS
|dw:1437609016552:dw|
DanJS
  • DanJS
The points are like the horizontal tangent on the graph of y=x^3, increasing both sides
anonymous
  • anonymous
I just dont get how you got the horizontal tangent?
DanJS
  • DanJS
you took the derivative, and found that sin(x) = 1 , when the derivative is zero right?
anonymous
  • anonymous
OH.. yea.. those values we got were the horizontal tangent ...
DanJS
  • DanJS
yes, when you hear derivative, think "slope of a tangent line there"
anonymous
  • anonymous
Yea, just forgot for a second there
anonymous
  • anonymous
So is function increasing from (-2pi, 2pi)?
DanJS
  • 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
DanJS
  • DanJS
right, increasing everywhere
DanJS
  • DanJS
with horizontal tangents at those two pooints you found
anonymous
  • anonymous
so for the horizontal tangents I got points: (-pi/2, - pi/2) and the (-3pi/2, -5.712)
anonymous
  • anonymous
and then for the second derivative, I would get: g''(t) = cos (t)
anonymous
  • anonymous
And for its critical values I would get pi/2 and -3pi/2, the same values we got before for the first derivative?
DanJS
  • DanJS
1 Attachment
DanJS
  • DanJS
Those are all the points you can add before the second derivative test
DanJS
  • DanJS
any up to there you dont understand
anonymous
  • 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)
anonymous
  • anonymous
For the inflection point: I got (-3pi/2, -3pi/2)
anonymous
  • anonymous
I have a good feeling im right
DanJS
  • DanJS
\[F(x) = x + \cos(x)\] \[F'(x) = 1 + \sin(x)\] \[F''(x) = -\cos(x) \]
DanJS
  • DanJS
second derivative is negative cosine
anonymous
  • anonymous
wait...but doesnt -sin = cos?
anonymous
  • anonymous
so that would just make it cos(t)
DanJS
  • DanJS
sorry back
anonymous
  • anonymous
lol k you almost gave me a heartattack
DanJS
  • DanJS
yes, so cos(x) = 0 at pi/2 and 3pi/2 plus and minus since you can move 2 pi backwards
anonymous
  • anonymous
we dont use -3pi/2? Only 3pi/2?
DanJS
  • DanJS
1 Attachment
DanJS
  • DanJS
notice where cos(x) is zero in the interval
anonymous
  • anonymous
wow, so do I use all those points?
DanJS
  • DanJS
Cos(x) = 0 when the angle is at 90 or 270 , or any addition or subtraction of 360
DanJS
  • DanJS
Cos(x) = 0 , x = pi/2 , x = 3pi/2, take off 2 pi from both x = -pi/2, x = -3pi/2
DanJS
  • DanJS
It is the X coordinate on the unit circle, zero on the Y axis at 90 and 270 degrees
anonymous
  • anonymous
yea makes sense
DanJS
  • DanJS
So those are your 4 test points
anonymous
  • anonymous
k thanks
DanJS
  • DanJS
welcome, the graph of the actual function is above for reference..

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