Question

the derivative of the function f is given by f'(x)=x^2cos(x^2). How many points of inflection does the graph of f have on the open interval (-2,2)

Answer 1

The point of inflection is calculated by equating the second derivative to zero and determining x from there.

f"(x) = -x²2xsinx² + cosx²(2x) = 0

2xcosx² - 2x³sinx² = 0
2x (cosx² - xsinx²) = 0

2x = 0   ⇒  x = 0
cosx² - xsinx² = 0   ⇒ x = 3.82 (if you use shift+solve in your scientific calculator)

Thus, the function only has 1 point of inflection and it is at x = 0.