Critical numbers help? I will reward with a medal!

- anonymous

Critical numbers help? I will reward with a medal!

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

\[\sin^2x+cosx\]
\[0

- zepdrix

Get that derivative? :U

- anonymous

sinx (2cosx -1)

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## More answers

- zepdrix

Mmm ok looks good! :)

- 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

sine is zero at what angles?

- anonymous

I'm not sure how to get that answer.

- zepdrix

Don't remember your trig..? :(

- 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

Is it a particular formula, or is it just something you'd have to memorize?

- zepdrix

Ugh OpenStudy freezing on me again.....

- anonymous

It does that to me a lot.

- anonymous

(this may seem like a stupid question, but which part is sine?

- zepdrix

|dw:1444434263248:dw|

- zepdrix

|dw:1444434276482:dw|sine is the y-coordinates

- zepdrix

So we want to know which angles give us a y-coordinate of 0

- anonymous

So the 180 angle

- zepdrix

Hmm there's another one :d

- anonymous

Oops! Didn't see that. The 0 angle

- zepdrix

Oh wait... was your interval really given like this?
\(\large\rm 0

- anonymous

That's what it looked like.

- 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

\[\large\rm 2\cos x-1=0\]Solving for cos x,\[\large\rm \cos x=\frac{1}{2}\]

- zepdrix

|dw:1444434611166:dw|Cosine is your x-coordinates

- anonymous

So 60 degrees for cosine?

- zepdrix

|dw:1444434707066:dw|

- anonymous

I'm not sure why I keep looking only in the first section. Thank you again for pointing that out.

- anonymous

60 and 300

- zepdrix

Degrees are icky :) lol
You gotta try to get comfortable with radians XD

- zepdrix

Ok great so we've found all of our critical points!!
\[\large\rm x=\frac{\pi}{3},~\pi,~\frac{5\pi}{3}\]

- 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

Wow! I never knew you could graph something like this on desmos. That's very interesting.

- anonymous

My connection on this site is horrible. sorry

- anonymous

Thank you for your help! (I'll probably be back on here soon with more questions)

- 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

Could you help me with one last thing really quickly?

- zepdrix

sure

- anonymous

I know the derivative is sinx (2cosx -1)
but I can't exactly figure out how to get that.

- zepdrix

oh you cheated on the first part? come on broski -_-

- anonymous

No, I didn't cheat.

- anonymous

The assignment gives me the derivative

- zepdrix

Oh :O

- 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

You really have to remember your sine and cosine derivatives to get through this one :)

- zepdrix

\[\large\rm y'=2(\sin x)(\cos x)+(-\sin x)\]

- zepdrix

And then factor a sin x out of each term.

- anonymous

Oh okay! I see now. Thank you. :)

- zepdrix

:D

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