Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t), [0, π/2]

Answer 1

y = 2cost + sin2t
differentiate wrt t
dy/dt = -2sint +2cos2t = 0
sint = cos2t = 1 - 2sin²t
2sin²t + sint -1 = 0
2sin²t + 2sint - sint - 1 = 0
2sint (sint +1 ) -(sint+1) = 0
sint = 1/2 and -1
but t€ [ 0, π/2 ] so, sint = 1/2 = sinπ/6
t = π/6
again, differentiate wrt t
d²y/dt² = -2cost -4sin2t
put t = π/6
d²y/dt² = -2× √3/2 - 4×√3/2 < 0
hence, d²y/dt² < 0
So, at t = π/6 function gain, Maxima

now,
f(0) = 2cos(0) +sin2(0) = 2
f(π/6) = 2cos(π/6) + sin2(π/6) = 3√3/2
f(π/2) = 2cos(0) + sin2(π/2) = 0

hence , absolute maximum = 3√3/2
and absolute minimum = 0