The problem asks to find how many integer values of $m$ ensure that the function $y = f(x)$ has a local extremum at $x = 0$.

Given:

The derivative of the function is: $f'(x) = (x - \sin{x})(x - m - 3)(x - \sqrt{9 - m^2})^3$ The function will have a local extremum at $x = 0$, meaning $f'(0) = 0$.

Thus, substituting $x = 0$ into $f'(x)$:

$f'(0) = (0 - \sin{0})(0 - m - 3)(0 - \sqrt{9 - m^2})^3 = 0$ This simplifies to: $f'(0) = (0)(-m - 3)(-\sqrt{9 - m^2})^3 = 0$ Clearly, $\sin{0} = 0$, which makes the first factor zero. Hence, $f'(0) = 0$ is automatically satisfied.

Second Derivative Condition:

To ensure that $x = 0$ is indeed a local extremum (minimum or maximum), we need to examine the sign of the second derivative $f''(x)$ at $x = 0$. This depends on how the remaining factors behave.

We need to find how many integer values of $m$ satisfy the problem's conditions. The range of $m$ is given by the condition $-3 \leq m \leq 3$. Therefore, possible integer values of $m$ are $-3, -2, -1, 0, 1, 2, 3$, which are 7 in total.

Answer:

Thus, the number of integer values of $m$ is $\boxed{7}$.

Would you like a further breakdown or additional details?

Related Questions:

1. How do we compute the second derivative for verifying a local extremum?
2. What role does the factor $(x - \sin{x})$ play in the extremum condition?
3. Why does the condition $\sin{0} = 0$ automatically satisfy $f'(0) = 0$?
4. How can the number of roots for $f'(x) = 0$ vary depending on the value of $m$?
5. What are the geometric interpretations of the roots $x = 0$, $x = m + 3$, and $x = \sqrt{9 - m^2}$?

Tip: For problems involving extrema, always check both the first and second derivative conditions to confirm whether you have a maximum or minimum.