anonymous
  • anonymous
Critical numbers help? I will reward with a medal!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sin^2x+cosx\] \[0
zepdrix
  • zepdrix
Get that derivative? :U
anonymous
  • anonymous
sinx (2cosx -1)

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zepdrix
  • zepdrix
Mmm ok looks good! :)
zepdrix
  • zepdrix
\[\large\rm \sin x (2\cos x -1)=0\]Apply your Zero-Factor Property\[\large\rm \sin x=0\qquad\qquad\qquad 2\cos x-1=0\]
zepdrix
  • zepdrix
sine is zero at what angles?
anonymous
  • anonymous
I'm not sure how to get that answer.
zepdrix
  • zepdrix
Don't remember your trig..? :(
anonymous
  • anonymous
I was horrible at memorizing trig stuff. I enjoyed solving certain types of problems with them, but quite frankly I don't remember much of anything except for a few identities.
anonymous
  • anonymous
Is it a particular formula, or is it just something you'd have to memorize?
zepdrix
  • zepdrix
Ugh OpenStudy freezing on me again.....
anonymous
  • anonymous
It does that to me a lot.
anonymous
  • anonymous
(this may seem like a stupid question, but which part is sine?
zepdrix
  • zepdrix
|dw:1444434263248:dw|
zepdrix
  • zepdrix
|dw:1444434276482:dw|sine is the y-coordinates
zepdrix
  • zepdrix
So we want to know which angles give us a y-coordinate of 0
anonymous
  • anonymous
So the 180 angle
zepdrix
  • zepdrix
Hmm there's another one :d
anonymous
  • anonymous
Oops! Didn't see that. The 0 angle
zepdrix
  • zepdrix
Oh wait... was your interval really given like this? \(\large\rm 0
anonymous
  • anonymous
That's what it looked like.
zepdrix
  • zepdrix
Oh ok :) Then I guess we can throw out the 0, since it's not in our interval. So then, \(\large\rm \sin x=0\) implies that our angle \(\large\rm x=\pi\). Ok great, we've found one critical point.
zepdrix
  • zepdrix
\[\large\rm 2\cos x-1=0\]Solving for cos x,\[\large\rm \cos x=\frac{1}{2}\]
zepdrix
  • zepdrix
|dw:1444434611166:dw|Cosine is your x-coordinates
anonymous
  • anonymous
So 60 degrees for cosine?
zepdrix
  • zepdrix
|dw:1444434707066:dw|
anonymous
  • anonymous
I'm not sure why I keep looking only in the first section. Thank you again for pointing that out.
anonymous
  • anonymous
60 and 300
zepdrix
  • zepdrix
Degrees are icky :) lol You gotta try to get comfortable with radians XD
zepdrix
  • zepdrix
Ok great so we've found all of our critical points!! \[\large\rm x=\frac{\pi}{3},~\pi,~\frac{5\pi}{3}\]
zepdrix
  • zepdrix
Here is the graph in case you wanted to see it, it's pretty cool https://www.desmos.com/calculator/6sbfzaocym You can clearly see a hill top at pi/3 and 5pi/3 and the bottom of a valley at the pi.
anonymous
  • anonymous
Wow! I never knew you could graph something like this on desmos. That's very interesting.
anonymous
  • anonymous
My connection on this site is horrible. sorry
anonymous
  • anonymous
Thank you for your help! (I'll probably be back on here soon with more questions)
zepdrix
  • zepdrix
Yah it's a little complicated to graph just a small interval on desmos, you have to put it into set brackets { } and the interval comes first, then colon, then your function.
anonymous
  • anonymous
Could you help me with one last thing really quickly?
zepdrix
  • zepdrix
sure
anonymous
  • anonymous
I know the derivative is sinx (2cosx -1) but I can't exactly figure out how to get that.
zepdrix
  • zepdrix
oh you cheated on the first part? come on broski -_-
anonymous
  • anonymous
No, I didn't cheat.
anonymous
  • anonymous
The assignment gives me the derivative
zepdrix
  • zepdrix
Oh :O
zepdrix
  • zepdrix
\[\large\rm y=(\sin x)^2+\cos x\]\[\large\rm y'=2(\sin x)(\sin x)'+(\cos x)'\]Start by applying power rule to the sine function. Then we have chain rule as well.
zepdrix
  • zepdrix
You really have to remember your sine and cosine derivatives to get through this one :)
zepdrix
  • zepdrix
\[\large\rm y'=2(\sin x)(\cos x)+(-\sin x)\]
zepdrix
  • zepdrix
And then factor a sin x out of each term.
anonymous
  • anonymous
Oh okay! I see now. Thank you. :)
zepdrix
  • zepdrix
:D

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