Find the absolute maximum and absolute minimum values of f on the given interval [0, pi/2]. f(t) = 2 cos t + sin 2t?? I got the minimum as 0 but can't find max??
Take the derivative and set it equal to 0. f(t) = 2 cos t + sin 2t f '(t) = -2 sin t + 2cos 2t 0 = -2 sin t + 2cos 2t 2 sin t = 2cos 2t sin t = cos 2t This happens at π/6 I believe and is probably either the max or the min but you also need to check the endpoints.
then i plug pi/6 into f(t)?
Check the derivative on either side of it: f '(π/9) = -2 sin π/9 + 2cos 2(π/9) which is + f '(π/4) = -2 sin (π/4) + 2cos (2(π/4)) which is - So my guess is that π/6 is a max
The min can be found by plugging 0 and π/2 into the original equation and seeing which one is smaller
\[f \prime (t)=2cos(2t)-2sin(t) \rightarrow sin(t)=cos(2t) \rightarrow t=arcsin(cos(2t)) \] when arcsin(1)=0, so we want cos(2t)=1. therefore x=0, is a critical point. since arcsin(x)=0 if x=1, so the general solution is cos(n*pi/2) where neZ. at Pi/2 we have another critical point f(0) and f(Pi/2) are your critical points. should find that f(0) is your max, not your min. good luck
Alex I would have to disagree because if you look at the graph of the original then it's pretty clear that π/6 is the max and π/2 is the min
Hmmm, you're right. Weird, can't see my mistake
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