$$f(x)=-5\sin (x)\cos (x)\text{ on }(-\pi ,\pi )$$
(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 $f^{\prime }(x)=-5({\cos }^2(x)+{\sin }^2(x))$.

Step 3 ：Simplify the derivative using the Pythagorean identity to get $f^{\prime }(x)=-5$.

Step 4 ：The critical numbers are the values of x for which $f^{\prime }(x)=0$. 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: $\boxed{{\displaystyle (-\pi ,\pi )}}$.

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