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#.
