Question

Answer the following questions for the function f(x)=sin^2(x/2) defined on the interval [-5.38, 1.47]. a.) f(x) is concave down on the region(s) 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(s)

Answer 1

The function f(x) = sin^2(x/2)is concave down on the interval [0,π], has a global minimum at x=−π, a local maximum at x=0, and is increasing on (−5.38,0).

a.) The concavity of a function is determined by the sign of its second derivative. The second derivative of f(x) can be calculated by taking the derivative of the first derivative. In this case,
`f''(x) = -(1)/(2) sin(x) + (1)/(4) sin^(2) ((x)/(2) )`. To find where f′′(x)<0 (indicating concavity down), we look for intervals where
`sin(x) > 2sin^(2)((x)/(2) )`. This happens when 0<x<π, so f(x) is concave down on the interval [0,π].

b.) To find the global minimum, we need to locate points where the first derivative f′(x)=0 or is undefined. The critical points occur when sin(x)=0, which is x=nπ for integer n. On the interval [−5.38,1.47], the global minimum occurs at x=−π because sin(−π)=0 and f(−π)=0.

c.) Local maxima occur where f′(x)=0 and f′′(x)<0. Considering the interval [−5.38,1.47], a local maximum occurs at x=0 since f′(0)=0 and f′′(0)<0.

d.) The function is increasing where, f′(x)>0. On the interval [−5.38,1.47], f′ (x)>0 for x<0 and f′ (x)<0 for x>0. Therefore, f(x) is increasing on the interval (−5.38,0)