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OpenStudy (anonymous):

Could some please help me with this? I have a test in a day and I just need some help so I can pass. How to find the critical values for y= cosx-sinx

OpenStudy (anonymous):

its 0213

Directrix (directrix):

My private message function on OS does not work. I suppose the "0" was truncated. Oh, well.

OpenStudy (anonymous):

you take the first derivative and set it equal to 0

OpenStudy (anonymous):

you take the first derivative and set it equal to 0

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I did (-sinx)=0 x=-sin^-1 (0) x=0, -pi, -2pi. For cos, i did (-cosx)=0 x=-cos^-1 (0) x= -pi/2, -3pi/2

OpenStudy (anonymous):

I did (-sinx)=0 x=-sin^-1 (0) x=0, -pi, -2pi. For cos, i did (-cosx)=0 x=-cos^-1 (0) x= -pi/2, -3pi/2

OpenStudy (anonymous):

Chat with you every night?!

OpenStudy (anonymous):

I can help out

OpenStudy (anonymous):

Alright I'll do what I can.

OpenStudy (anonymous):

I have skype

OpenStudy (anonymous):

Hmmm, I don't use that.

OpenStudy (anonymous):

I don't really use that either. Skype is the only thing I use daily. Sorry

OpenStudy (anonymous):

I think I can get it

Directrix (directrix):

@0213 Some woman came on this site and told me that I should go to www.tutor.com for one on one help. But, that comes with a price tag. Also, Thompson has a tutoring site. Let me look for the URL of that site.

Directrix (directrix):

Cry and let out the stress, I say. ------------------------ http://www.freewebs.com/jimthompson5910/home.html This is Thompson's site. He is also on Twitter: https://twitter.com/jimthompson5910 jim_thompson5910@hotmail.com Did @wio answer this question: How to find the critical values for y= cosx-sinx

OpenStudy (anonymous):

I can help with the question itself.

OpenStudy (anonymous):

thank you.

OpenStudy (anonymous):

Want me to walk you through it right now?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

First we find \(y'\). In this case: \[ y'= (\cos x-\sin x)' = -\sin x - \cos x \]So we want to find the roots of \( -\sin x - \cos x\) to get critical numbers.

OpenStudy (anonymous):

Does this step make sense?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[ -\sin x -\cos x = 0 \implies -\cos x = \sin x \implies -1 = \frac{\sin x}{\cos x}=\tan x \]

OpenStudy (anonymous):

So basically, we want to know when \[ \tan x = -1 \]Did you follow that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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OpenStudy (anonymous):

So there are two angles in the first revolution. \(90^\circ + 45^\circ = 135^\circ\) \(-90^\circ + 45^\circ = -45^\circ\)

OpenStudy (anonymous):

Did you follow that at all?

OpenStudy (anonymous):

I think. Since you want to know what tanx=-1 is. and tan is sin/cos, then you use the cast rule to find the values of when tan will be -1. since its negative , you find it were the value will be negative in the cast rule

OpenStudy (anonymous):

Basically, we know that \(|\cos x| = |\sin x|\) at the \(45^\circ\) angle parts... So when we want \(\cos x = -\sin x\) we make them equal but only in certain quadrants.

OpenStudy (anonymous):

Anyway, in radians, our solutions for the first rotation is \[ \frac{3\pi}{4}, -\frac{\pi}{4} \]

OpenStudy (anonymous):

To get all solutions, we add on full rotations to the angles. Where \(n \in \mathbb{Z}\) (meaning \(n\) is whole number) \[ \frac{3\pi}{4} + 2\pi n, -\frac{\pi}{4} + 2\pi n \]

OpenStudy (anonymous):

Remember that any angle \(+ 2\pi\) is going to be that same angle because \(2\pi\) is a full rotation.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

thank you so much @wio for your help.

OpenStudy (anonymous):

I can try to help you with problems if you use Skype. Calculus is very important for any STEM degree, and STEM degrees are the most valuable degrees today.

Directrix (directrix):

@0213 It is too soon to decide whether you should stop studying Calculus. Everybody hits a bump during the first year of University. It is the transition from high school to University. I would not listen to any person in my class who said bad things about my work. Have you seen their grades? They may be failing themselves. Find one good friend and don't talk to those others. Sit on the front row of the classroom. Get some help (does your professor have office hours?) from somebody you can understand. I don't care for Skype myself but it would make explaining stuff easier, I think. You need to get some rest, a good meal, and some exercise. Then, come back to Calculus and make a plan on how to get your questions answered. It seems to me that you know the gist of the Calculus you are studying but have some points of confusion that need to be cleared up.

OpenStudy (anonymous):

The only advantage I see for skype is that I'm on it more than OS so it'd be another chance to find me. Also I could answer short questions on it too.

OpenStudy (anonymous):

How do I add you on skype @wio

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