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