Question

f(x) = sin^2(x/2)
defined on the interval [ -5.683185, 1.270796].

Remember that you can enter pi for \pi as part of your answer.

a.) f(x) is concave down on the interval .

b.) A global minimum for this function occurs at .

c.) A local maximum for this function which is not a global
maximum occurs at .

d.) The function is increasing on the region .

Answer 1

f = sin^2(x/2)
f' = 2sin(x/2)cos(x/2)(1/2) = sin(x)/2
f'' = cos(x)/2

concave down where f'' < 0: [-3pi/2,-pi/2]

global min where f'=0, f''>0: 0

local max where f'=0, f''<0: -pi
This is also the global max

f is increasing where f'>0: [-5.683185,-pi)U(0,1.270796]

Answer 2

Technically,
concave down where f'' < 0: (-3pi/2,-pi/2)

Answer 3

thank you, that really help me