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\[f(x)= \sin (x)+\cos (x), 0\le x \le2\pi\]
i guess what i really need help with is the critical points in the first derivative text and in the second derivative test.
\[f'(x) = \cos(x) - \sin(x) = Rcos(x + \alpha)\] 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
hmm thanks i'm pretty sure i need to approach that differently
You think I'm wrong?
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
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
Well, my method gets you the critical values of d/dx f(x). And your method doesn't exist.
Maybe you should ask Leibniz instead ;-)
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
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
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:
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.
You could use the same method on the original equation (not the derivative) - if you are good at sketching, it's quicker.