\[ f(x)=-5 \sin (x) \cos (x) \text { on }(-\pi, \pi) \] (a) Find the critical numbers of $f$. (Separate multiple answers by commas.) (b) Determine the open intervais on which $f$ is increasing and decreasing. $f$ is increasing on: $f$ is decreasing on: (c) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. Relative maxima oceur at $x=$ (Separate multiple answers by commas.) Relative minima occur at $x=$ (Separate multiple answers by commas.)

Step 1 ：Find the derivative of the function \(f(x) = -5 \sin(x) \cos(x)\) using the product rule and chain rule.

Step 2 ：The derivative is \(f'(x) = -5(\cos^2(x) + \sin^2(x))\).

Step 3 ：Simplify the derivative using the Pythagorean identity to get \(f'(x) = -5\).

Step 4 ：The critical numbers are the values of x for which \(f'(x) = 0\). However, the derivative of the function is a constant and does not equal to zero for any x. Therefore, the function has no critical numbers.

Step 5 ：Since the derivative of the function is a negative constant, the function is decreasing on its entire domain, which is \((-\pi, \pi)\).

Step 6 ：So, \(f\) is decreasing on: \(\boxed{(-\pi, \pi)}\).

Step 7 ：Since there are no critical numbers, there are no relative maxima or minima.