Repeat Exercise 23 for the interval $[-3,9]$

Key Concepts: Interval Notation, Function Properties Explanation: The problem asks us to perform the same steps as in Exercise 23 but for a different interval. We will find the critical points and test for interval sign changes to find the intervals where the function is increasing/decreasing. We will then use the First Derivative Test to locate the extrema. Finally, we will determine the concavity and points of inflection by finding the second derivative, testing for intervals of sign change and using the Second Derivative Test. Step by Step Solution: Step 1. Find the first and second derivative of the function. Step 2. Determine the critical points of the function by setting the first derivative equal to zero and solving for x. The critical points are: x=-2 and x=4 Step 3. Test for the sign of the first derivative for each interval outside the critical points to determine if the function is increasing or decreasing. The intervals are: $(-\infty ,-2)$, $(-2,4)$, $(4,\infty )$. Since $f^{\prime }(x)=-3x^2+12x$ the sign of f'(x) is positive for $x<0$, negative for $0<x<4$, and positive for $x>4$. Therefore, the function is increasing on $(-\infty ,-2)$ and $(4,\infty )$ and decreasing on $(-2,4)$ Step 4. Locate the extrema by using the First Derivative Test. We have a local maximum at $x=-2$ and a local minimum at $x=4$. Step 5. Determine the intervals of convexity and concavity by finding the second derivative of the function. The second derivative is $f^{\prime \prime }(x)=-6x+12$. Solving $f^{\prime \prime }(x)=0$ yields the inflection point $x=2$. Testing the intervals determined in Step 3 yields that the function is concave down on $(-\infty ,2)$ and concave up on $(2,\infty )$. Therefore, the point of inflection is $(2,-4)$. Step 6. Finally, sketch the graph and label the key points: extrema and inflection point