Question

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??

Answer 1

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.

Answer 2

then i plug pi/6 into f(t)?

Answer 3

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

Answer 4

The min can be found by plugging 0 and π/2 into the original equation and seeing which one is smaller

Answer 5

\[f \prime (t)=2cos(2t)-2sin(t) \rightarrow sin(t)=cos(2t) \rightarrow t=arcsin(cos(2t)) \]
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

Answer 6

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

Answer 7

Hmmm, you're right. Weird, can't see my mistake