The problem is asking for the values of $x$ where the function $f(x) = x^3 - 3x^2 - 15$ is decreasing.

To solve this, we first need to find the derivative of $f(x)$ and analyze where it is negative (since a function decreases when its derivative is negative).

Step 1: Derive $f(x)$

The function is: $f(x) = x^3 - 3x^2 - 15$ We find the first derivative $f'(x)$: $f'(x) = 3x^2 - 6x$

Step 2: Factor the derivative

Now, factor the derivative expression: $f'(x) = 3x(x - 2)$

Step 3: Determine where $f'(x)$ is negative

To find where $f'(x)$ is negative, analyze the critical points by solving $f'(x) = 0$: $3x(x - 2) = 0$ This gives the critical points $x = 0$ and $x = 2$.

Next, we use these critical points to determine the intervals where $f'(x)$ is negative. Test the sign of $f'(x)$ in the following intervals:

• For $x < 0$ (e.g., $x = -1$): $f'(-1) = 3(-1)((-1) - 2) = 3(-1)(-3) = 9 > 0$ So, $f'(x) > 0$ in this interval (function is increasing).

• For $0 < x < 2$ (e.g., $x = 1$): $f'(1) = 3(1)(1 - 2) = 3(1)(-1) = -3 < 0$ So, $f'(x) < 0$ in this interval (function is decreasing).

• For $x > 2$ (e.g., $x = 3$): $f'(3) = 3(3)(3 - 2) = 3(3)(1) = 9 > 0$ So, $f'(x) > 0$ in this interval (function is increasing).

Step 4: Conclusion

The function decreases when $0 < x < 2$. Thus, the correct answer is:

B. $0 < x < 2$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you find the intervals where a function is increasing or decreasing?
2. What is the significance of the derivative being zero at critical points?
3. Can a function have both increasing and decreasing intervals? How?
4. How would you determine the concavity of a function using the second derivative?
5. What role do critical points play in sketching the graph of a function?

Tip: Always test the sign of the derivative in each interval to determine whether a function is increasing or decreasing.