Given f(x) = 2 cos(x) + sin(2x), we use the first derivative test to get
f'(x) = -2sin(x) + 2cos(2x) = 0
= -sin(x) + 1 - 2sin^2(x) = 0
Let t = sin(x) then
-2t^2 - t + 1 = 0 or 2t^2 + t - 1 = 0
this factors to (t + 1)(2t - 1) so that t = -1 and t = 1/2
Since t = sin(x), we get
-1 = sin(x) and 1/2 = sin(x)
This mean the first derivative is equal to zero at x = 3pi/2 and pi/6. Since 3pi/2 is outside the interval, we can ignore this.
We find the second derivative to see if the function at pi/6 is a max or min.
f''(x) = -cos(x) - 4sin(2x)
Plugging in pi/6, we get f''(pi/6) = -cos(pi/6) - 4sin(pi/3) = -0.8660 - 4(0.8660) < 0 so this is a maximum.
To make sure we find the absolute maximum, we plug in the endpoints into the original equation so we get
f(0) = 2cos(0) + sin(0) = 2
f(pi/2) = 2cos(pi/2) + sin(pi) = 0
Plugging in the x value pi/6 for the maximum we found above we get
f(pi/6) = 2cos(pi/6) + sin(pi/3) = 1.732 + 0.8660 = 2.56 which is the ABSOLUTE maximum in the interval [0,pi/2] since it's the largest...