Find any critical numbers in [0, pi/2],
then find the value of f at 0, the critical numbers in [0, pi/2], and at pi/2.
f(t) = 2cost + sin2t
f'(t) = -2sint + 2cos2t
= -2(sint - (1-2sin^2t)
= -2(2sin^2 t+sint-1)
= -2(2sint-1)(sint+1).
So f'(t) is never undefined and is 0 where:
sint = 1/2 " " or " " sint = -1.
The only solution in the interval [0, pi/2] is t = pi/6.
f(0) = 2cos(0) + sin2(0) = 2 .
f(pi/6) = 2cos(pi/6) + sin(pi/3) = (2sqrt3)/2 + sqrt3 = (3sqrt3)/2.
f(pi/2) = 2cos(pi/2) + sinpi = 0 .
The minimum is 0 (and it occurs at pi/2).
The maximum is (3sqrt3)/2 (and it occurs at pi/6).
Note:
We can see that (3sqrt3)/2 > 2 by using the fact that, for numbers greater than 1, the square of the greater number is greater.
The square of (3sqrt3)/2 is (9*3)/4 = 27/4
while the square of 2 is 4 = 16/4.
So (3sqrt3)/2 > 2.