Given the function f (x) = 3cos x for the interval 0 < x < 2pi, determine the concavity of the function, between the critical values

Answer 1

f'(x) = -3sinx
f'=0 at x = 0, π and 2π

f"(x) = -3cosx
f"(0) < 0 so f is concave down in (0,π/2)
f"(π) > 0 so f is concave up in (π/2,3π/2)
f"(2π) < 0 so f is concave down in (3π/2,2π)