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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

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

try -2pi

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 ,

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

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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)?

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)

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

any up to there you dont understand

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)

sorry back

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

k thanks

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