Final Answer:
On the interval [0, π], the function f(x) = sin(x) has critical points at x = 0 and x = π.
Step-by-step explanation:
To determine critical points of the function f(x) = sin(x) on the interval [0, π], we first find the derivative f'(x). The derivative of sin(x) is f'(x) = cos(x). Critical points occur where the derivative is either zero or undefined. However, on the interval [0, π], the derivative cos(x) is never undefined. To find where the derivative is zero, solve the equation cos(x) = 0.
In the interval [0, π], cos(x) = 0 at x = π/2 and x = 3π/2. However, x = 3π/2 is not in the interval [0, π]. Hence, the only critical point of f(x) = sin(x) on [0, π] is at x = π/2. This point corresponds to a relative maximum because the function changes from increasing to decreasing at that point.
Therefore, after analyzing the first derivative, we find that the critical point of f(x) = sin(x) on the interval [0, π] occurs at x = π/2, while there are no critical points at x = 0 within this interval as the function does not reach a local extremum there