Answer:
maximum: 8√3/9 ≈ 1.5396
minimum: -8√3/9 ≈ -1.6596
Step-by-step explanation:
Extreme values are found where the derivative of the function is zero. Here, the zeros of the derivative are more easily found if the function is written in terms of sin(x).
. . . f(t) = 2cos(t)sin(2t) = 2cos(t)·(2sin(t)cos(t)) = 4cos(t)²sin(t) = 4(1 -sin(t)²)sin(t)
. . . f(t) = 4sin(t) -4sin(t)³
The derivative is ...
. . . f'(t) = 4cos(t) -4(3sin(t)²cos(t)) = 4cos(t)(1 -3sin(t)²)
This has zeros at odd multiples of pi/2, and where sin(t) = ±√(1/3). Using sin(t) = ±√(1/3) in f(t), we have ...
. . . f(t) = (±4/√3)(1 -1/3) = ±(8/3)/√3 = ±(8/9)√3
The extreme values of f(t) are ±(8/9)√3, approximately ±1.5396.