Answer the following questions about the function whose derivative is $f^{\prime}(x)=x(x+3)$.
a. What are the critical points of $f$ ?
b. On what open intervals is $f$ increasing or decreasing?
c. At what points, if any, does $f$ assume local maximum and minimum values?
a. Find the critical points, if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The critical point(s) of $f$ is/are $x=-3,0$
(Simplify your answer. Use a comma to separate answers as needed.)
B. The function $f$ has no critical points.
b. Determine where $\mathrm{f}$ is increasing and decreasing. Select the cerrect choice below and fill in the answer box to complete your choice.
(Type your answer in interval notation. Use a comma to separate answers as needed.)
A. The function $\mathrm{f}$ is increasing on the open interval(s) $\square$, and never decreasing.
B. The function $\mathrm{f}$ is decreasing on the open interval(s) $\square$, and never increasing.
C. The function $\mathrm{f}$ is increasing on the open interval(s) $\square$, and decreasing on the open interval(s) $\square$

Step 1 ：The critical points of a function are the points where the derivative is either zero or undefined. In this case, the derivative of the function is \(f'(x) = x(x+3)\). To find the critical points, we set the derivative equal to zero and solve for x: \(x(x+3) = 0\). This gives us two solutions: \(x = 0\) and \(x = -3\). Therefore, the critical points of the function are \(x = 0\) and \(x = -3\).

Step 2 ：To determine where the function is increasing or decreasing, we need to examine the sign of the derivative on the intervals determined by the critical points. The critical points divide the number line into three intervals: \((-\infty, -3)\), \((-3, 0)\), and \((0, \infty)\).

Step 3 ：For the interval \((-\infty, -3)\), we can choose \(x = -4\). Substituting this into the derivative gives us \((-4)(-4+3) = -4\), which is negative. Therefore, the function is decreasing on the interval \((-\infty, -3)\).

Step 4 ：For the interval \((-3, 0)\), we can choose \(x = -1\). Substituting this into the derivative gives us \((-1)(-1+3) = 2\), which is positive. Therefore, the function is increasing on the interval \((-3, 0)\).

Step 5 ：For the interval \((0, \infty)\), we can choose \(x = 1\). Substituting this into the derivative gives us \((1)(1+3) = 4\), which is positive. Therefore, the function is increasing on the interval \((0, \infty)\).

Step 6 ：The function assumes local maximum and minimum values at the critical points. Since the function changes from decreasing to increasing at \(x = -3\), there is a local minimum at \(x = -3\). Since the function changes from increasing to increasing at \(x = 0\), there is no local maximum or minimum at \(x = 0\).

Step 7 ：Therefore, the function has a local minimum at \(x = -3\) and no local maximum. \(\boxed{\text{Local minimum at } x = -3}\)