F(t) = 2 cos t + sin 2t, [0, π/2]

find absolute max and min

f(t) = 2cos(t) + sin(2t) [0, pi/2]

The absolute min/max is found by evaluating the function at
1) the endpoints, and
2) local minima/maxima

Let's start with the endpoints.
f(0) = 2cos(0) + sin(2*0) = 2(1) + 0 = 2
f(pi/2) = 2cos(pi/2) + sin(2*pi/2)
f(pi/2) = 2(0) + sin(pi) = 0 + 0 = 0

Now, let's calculate the local minima/maxima. Take the derivative.

f'(t) = 2(-sin(t)) + cos(2t) (2)
f'(t) = -2sin(t) + 2cos(2t)

Make f'(t) = 0,

0 = -2sin(t) + 2cos(2t). Use the double angle formula.
0 = -2sin(t) + 2[cos^2(t) - sin^2(t)]
0 = -2sin(t) + 2cos^2(t) - 2sin^2(t)
0 = -2sin(t) + 2[1 - sin^2(t)] - 2sin^2(t)
0 = -2sin(t) + 2 - 2sin^2(t) - 2sin^2(t)
0 = -2sin(t) + 2 - 4sin^2(t)
0 = -sin(t) + 1 - 2sin^2(t)
2sin^2(t) + sin(t) - 1 = 0
[2sin(t) - 1] [sin(t) + 1] = 0

Obtain all solutions within [0, pi/2].

2sin(t) - 1 = 0
2sin(t) = 1
sin(t) = 1/2
t = { pi/6, 5pi/6 } but 5pi/6 is outside of the interval, so reject it.

sin(t) + 1 = 0
sin(t) = -1
t = {3pi/2} but reject this because it's outside of our interval.

Therefore, critical numbers are only
t = pi/6

Plug this value into the function.

f(pi/6) = 2cos(pi/6) + sin(2*pi/6)
f(pi/6) = 2[sqrt(3)/2] + sin(pi/3)
f(pi/6) = sqrt(3) + sqrt(3)/2
f(pi/6) = [2sqrt(3) + sqrt(3)]/2
f(pi/6) = [3sqrt(3)]/2


Compare all values; endpoints and local extrema.

f(0) = 2
f(pi/2) = 0
f(pi/6) = [3sqrt(3)]/2 =~ 2.59 (approx)

The highest value among these values is the absolute maximum.
The lowest value among these values is the absolute minimum.

Absolute maximum: 3sqrt(3)/2
Absolute minimum: 0