Calculus 1: f(x) = ln(2 + sin(x)), 0 ≤ x ≤ 2𝜋

Question

f(x) = ln(2 + sin(x)), 0 ≤ x ≤ 2𝜋

Find the interval(s) on which f is concave up. (Enter your answer using interval notation.)

Find the interval(s) on which f is concave down. (Enter your answer using interval notation.)

Find the inflection points of f.

smaller x-value (x, y) = ( , )

larger x-value (x, y) = ( , )

Answer 1

desmos.com/calculator/tdqqwxogds

Answer 2

f(x) = ln(2 + sinx)

f'(x) = cosx / (2 + sinx)

f"(x) = [(2 + sinx)(-sinx) - cosx(cosx)] / (2 + sinx)² = [2 - (sin²x + cos²x) / (2 + sinx)² = 1 / (2 + sinx)²

f"(x) > 0 for all x. So, f is concave up for all x.

There are no inflection points. .