How do you find the global extreme values for f(t) = 2cost + sin2t on [0,pi/2]?

1 Answer
Aug 27, 2015

Use the closed interval method.

Explanation:

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.