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. \(\boxed{\text{Local minimum at } x = -3}\)