Question

i have to find the open intervals on which the function is increasing or decreasing. The function is: sin x+cos x. Can someone help me?

Answer 1

Okay, well, the first thing you want to do is differentiate it:
\[y'=\cos(x)-\sin(x)\]
Then set it to zero to find the maxes/mins.
\[0=\cos(x)-\sin(x) \rightarrow \sin(x)=\cos(x) \rightarrow \tan(x)=1\]
We know that tan(x)=1 at:
\[x=\frac{\pi}{4} \pm \pi\]
So, assuming you're in the interval:\[x \in [0,2 \pi]\] (as most trig problems are)
You have a zero at pi/4 and 3pi/4.
Now take a second derivative to test whether they are concave up or down. (max/min)
\[y''=-\sin(x)-\cos(x)\]
\[y''|_{x=\frac{\pi}{4}}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}<0\]
So this is concave DOWN or a relative MAX.
Then:
\[y''|_{x=\frac{5\pi}{4}}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}>0\]
So this is concave UP or a relative MIN.
This tells you that, starting from zero, it is increasing until it hits pi/4, then decreases from pi/4 to 5pi/4, the increases to 2pi.
So you would have:
increasing:
\[x \in (0,\frac{\pi}{4}) \cup (\frac{5\pi}{4},2 \pi)\]
decreasing:
\[x \in (\frac{\pi}{4},\frac{5\pi}{4})\]
And a wolfram link so you can look at the graph to assure yourself this is correct:
http://www.wolframalpha.com/input/?i=y%3Dsin(x)%2Bcos(x)

Answer 2

That should read: "You have a zero at pi/4 and 5pi/4". Just a mistype.

Answer 3

Thank you

Answer 4

No problem :P Just tell me if that confuses you or if there is something else you're having problems with :P

Answer 5

And this might be a slightly better picture because you can see the max and min points, and the decreasing between them. http://www.wolframalpha.com/input/?i=plot+y%3Dsin(x)%2Bcos(x)%2Cx%2Cpi%2F4%2C5pi%2F4

Answer 6

I have another question, if i am given the interval (0,2pi), how can i determine what angles go as critical points. Lets say, i found that sinx=1/2. What are the criticalpoints?

Answer 7

Okay, well:
\[\sin(x)=\frac{1}{2} \iff x=\frac{\pi}{6},\frac{5\pi}{6}\] on the interval (0,2pi). Those are just "unit circle" nice angles.
http://etc.usf.edu/clipart/43200/43215/unit-circle7_43215_lg.gif

Answer 8

Cos(x) corresponds to the x values, and sin(x) corresponds to the y values. You just have to watch the quadrant because cos(x) is negative in quadrants 2 and 3 where sin is negative in quadrants 3 and 4.

Answer 9

okay, thanks

Answer 10

No problem :P

Answer 11

one more question, how do i dtermine within the intervals whether the derivative is positive or not? Like for x/2+cos, i found the derivative to be 1/2-sinx. Then the crtical numbers are pi/6 and 5pi/6. The interval i am working within is (0,2pi). Then i found the open intervals as being (0,pi/6) (pi/6,5pi/6) (5pi/6,2pi). What value to i pick within each interval and how do i know whter the derivative will be positive or negative?

Answer 12

If you take the second derivative, it tells you concavity, so you know, if you plug in the critical point, and its POSITIVE, that you have something concave UP meaning its a relative MIN. This tells you that before it, it was decreasing, then after the point, it is INCREASING. Alternatively, you can use the first derivative test and plug in a number slightly bigger than the critical point to test, but the second derivative test is easier when you are dealing with simple differentiation. Does that make sense?

Answer 13

could you show me with the example i gave how to see if the first derivative is positive or negaivtive depending on the number we choose with the interval? I am having a hard time doing the first derivative test with trig functions. I am just trying to find out how i would know that the first derivative is neg or pos.

Answer 14

Okay. Well, if the first derivative is positive you know that the function is increasing, and if it is negative that it is decreasing (as the slope of the tangent). So, for example, if you had y'=cos(x) on the interval (0,2pi).
You know that cos(x) is zero at pi/2 and 3pi/2.
So those are our critical points. If you want to use the FIRST derivative test, simply plug in something such as pi/4,pi,and 7pi/4. You can use these numbers (even though they seem a good distance away) because, as long as they are between the critical points, they will work. This may seem counter intuitive, but if we found the critical points, anywhere in the middle obviously isn't a critical point. So, plugging in pi/4 you would get sqrt(2)/2 > 0 so the function is INCREASING. Then if you plug in pi you get -1<0 so the function is DECREASING. Then if you plug in 7pi/4 you get sqrt(2)/2 >0 so the function is INCREASING. This tells you that pi/2 is a MAXIMUM because it goes from increasing to decreasing, and that 3pi/2 is a MINIMUM because it goes from decreasing to increasing. Does this answer your question a little better? :P

Answer 15

yes thanks you

Answer 16

No problem xP

Answer 17

Why would i plugin -1

Answer 18

You wouldn't, you'd plug in pi, and the cos(pi)=-1

Answer 19

And the values that I picked aren't the only ones you could have used. You could have used pi/6 or pi/3 or some other combination. As long as the numbers you pick are before and after each critical point.