(1 point) If $f(x)=x^{2} \sqrt{25-x}$, find the open intervals on which $f$ is increasing and decreasing, and the coordinates of all extrema of $f$. $f(x)$ is increasing on $f(x)$ is decreasing on (Enter your answers using interval notation.) $f(x)$ has relative maxima at $f(x)$ has relative minima at $f(x)$ has absolute maxima at $f(x)$ has absolute minima at (Enter each extremum as an ordered pair, e.g., $(2,3)$. If there is more than one extremum of a given type, enter a comma-separated list of ordered pairs. If there are no extrema of a given type, enter "none".)

Step 1 ：Find the derivative of the function using the product rule and the chain rule: \(f'(x) = 2x\sqrt{25-x} + x^2(-1/2\sqrt{25-x})\)

Step 2 ：Simplify the derivative: \(f'(x) = 2x\sqrt{25-x} - \frac{x^2}{2\sqrt{25-x}}\)

Step 3 ：Set the derivative equal to zero and solve for x to find the critical points: \(2x\sqrt{25-x} - \frac{x^2}{2\sqrt{25-x}} = 0\)

Step 4 ：This equation simplifies to: \(4x^2(25-x) - x^4 = 0\)

Step 5 ：Solving this equation gives us the critical points x = 0, x = 5, and x = -5. We can ignore x = -5 because it is not in the domain of the original function.

Step 6 ：Test the intervals between the critical points in the derivative to determine where the function is increasing or decreasing. Choose test points -1, 3, and 6.

Step 7 ：Calculate \(f'(-1) = 2(-1)\sqrt{25-(-1)} - \frac{(-1)^2}{2\sqrt{25-(-1)}} = -2\sqrt{26} - \frac{1}{2\sqrt{26}} < 0\)

Step 8 ：Calculate \(f'(3) = 2(3)\sqrt{25-(3)} - \frac{(3)^2}{2\sqrt{25-(3)}} = 6\sqrt{22} - \frac{9}{2\sqrt{22}} > 0\)

Step 9 ：Calculate \(f'(6) = 2(6)\sqrt{25-(6)} - \frac{(6)^2}{2\sqrt{25-(6)}} = 12\sqrt{19} - \frac{36}{2\sqrt{19}} < 0\)

Step 10 ：So, the function is decreasing on the interval (-∞, 0), increasing on the interval (0, 5), and decreasing on the interval (5, ∞).

Step 11 ：The function has a relative minimum at (0, 0) and a relative maximum at (5, 0).

Step 12 ：Since the function is decreasing to the left of x = 0 and to the right of x = 5, and increasing between x = 0 and x = 5, the function has an absolute minimum at (0, 0) and an absolute maximum at (5, 0).

Step 13 ：\(\boxed{f(x)\text{ is increasing on }(0, 5)}\)

Step 14 ：\(\boxed{f(x)\text{ is decreasing on }(-∞, 0)\text{ and }(5, ∞)}\)

Step 15 ：\(\boxed{f(x)\text{ has relative maxima at }(5, 0)}\)

Step 16 ：\(\boxed{f(x)\text{ has relative minima at }(0, 0)}\)

Step 17 ：\(\boxed{f(x)\text{ has absolute maxima at }(5, 0)}\)

Step 18 ：\(\boxed{f(x)\text{ has absolute minima at }(0, 0)}\)