## anonymous 5 years ago A) find the intervals on which f is increasing or decreasing. B) find the local maximum and minimum values of f C) Find the intervals of concavity and the inflection points. Where....

1. anonymous

$f(x)= \sin (x)+\cos (x), 0\le x \le2\pi$

2. anonymous

i guess what i really need help with is the critical points in the first derivative text and in the second derivative test.

3. anonymous

$f'(x) = \cos(x) - \sin(x) = Rcos(x + \alpha)$ for some alpha. Expanding: cos(x) - sin(x) = Rcos(a)cos(x) - Rsin(a)(sin(x) => R cos(a) = R sin (a) = 1 => tan a = 1 => a = pi/4. R^2(sin^2(a) + cos^2(a)) = 2 => R = sqrt(2) Gogogogogog

4. anonymous

hmm thanks i'm pretty sure i need to approach that differently

5. anonymous

You think I'm wrong?

6. anonymous

f'(x) = 0 iff cos(x-pi/4) = 0 iff x-pi/4 = pi/2 , 3pi/2 iff x = 3pi/4, 7pi/4 You have your critical values

7. anonymous

the alpha and r is was confused me. i've never had to use that in calculus sorry. didnt mean to like offend or anything

8. anonymous

Well, my method gets you the critical values of d/dx f(x). And your method doesn't exist.

9. nowhereman

10. anonymous

Just realised I should take off pi/4 - sorry, I'm getting old. It was a copying error, rather than a mathematical one (because I don't make them). Critical values are pi/4, 5pi/4 Oh, and on the subject of your precious leibniz : http://imgs.xkcd.com/comics/newton_and_leibniz.png

11. anonymous

ha that was funny. i still dont understand where the R or alpha come from. i'm sorry i havent had trig in a really long time so these trig equations always throw me off

12. anonymous

In general, you can rewrite m * cos(x) + n sin(x) as R cos (x ± a) (or R sin (x ± a) because of how they expand. if you work out R and, more importantly for this problem, a, you can use it to work out the minimum/maximum/critical etc values. It's done because it's easier to work with an equation with once trig function than 2:

13. anonymous

There is probably some other way to do this, but I can't think of it right now, so it's probably not too good.

14. anonymous

You could use the same method on the original equation (not the derivative) - if you are good at sketching, it's quicker.