Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

Question

f(x)=sinx/2 , [0, 4π]

For inflection point, I got 0,0, but it was marked incorrect.

Concave up = (2pi,4pi) marked correct

Concave down (0, pi) marked incorrect.

Help?

Answer 1

Derivative f'(x) = 1/2cosx/2 and second derivative f''(x) = -1/4sinx/2;

To get inflection point let set equation f''(x) = 0: -1/4sinx/2 = 0; sin x/2 = 0; x/2 = 0, x/2 = π, or x/2 = 2π;

x = 0, x = 2π, x = 4π.

0 and 4π are end points of interval. Because only x = 2π can be x-coordinate of inflection point;

-1/4sin(x/2) < 0 for interval (0, 2π), f(x) graph concave down and -1/4sin(x/2) > 0 for interval (2π, 4π), here

graph f(x) concave up. If x = 2π f(2π) = sin(2π/2) = sinπ = 0.

Because (2π, 0) point of inflection