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Let's analyze the problem step by step.

We are given the function:

\[
f(x) = \frac{x^3 + 4x^2 + x - 6}{3 \sin\left(-\frac{\pi}{2} x\right) + 3x^2}
\]

We are tasked with determining why \( f \) is not continuous at \( x = 1 \).

### Step 1: Definition of Continuity

A function \( f(x) \) is continuous at \( x = 1 \) if:
1. \( \lim_{x \to 1} f(x) \) exists.
2. \( f(1) \) is defined.
3. \( \lim_{x \to 1} f(x) = f(1) \).

If any of these conditions is violated, the function is not continuous at \( x = 1 \).

### Step 2: Analyzing the function

We need to analyze two aspects:
1. **The limit** \( \lim_{x \to 1} f(x) \).
2. **The value of the function** \( f(1) \).

#### Finding \( f(1) \)
Substituting \( x = 1 \) into the expression for \( f(x) \):

\[
f(1) = \frac{1^3 + 4(1)^2 + 1 - 6}{3 \sin\left(-\frac{\pi}{2}(1)\right) + 3(1)^2} = \frac{1 + 4 + 1 - 6}{3 \sin\left(-\frac{\pi}{2}\right) + 3}
\]

\[
f(1) = \frac{0}{3 \cdot (-1) + 3} = \frac{0}{0}
\]

Since we end up with \( \frac{0}{0} \), this is indeterminate, meaning \( f(1) \) is **undefined**.

#### Finding \( \lim_{x \to 1} f(x) \)
Next, we must check if the limit exists. Because the numerator and denominator both tend to zero when \( x = 1 \), we can use **L'Hopital's Rule** to evaluate the limit:

Applying L'Hopital's Rule, we differentiate the numerator and denominator with respect to \( x \):

\[
\text{Numerator: } \frac{d}{dx}(x^3 + 4x^2 + x - 6) = 3x^2 + 8x + 1
\]

\[
\text{Denominator: } \frac{d}{dx}(3 \sin\left(-\frac{\pi}{2} x\right) + 3x^2) = 3 \left(-\frac{\pi}{2} \cos\left(-\frac{\pi}{2} x\right)\right) + 6x
\]

Now, we evaluate the limit:

\[
\lim_{x \to 1} \frac{3x^2 + 8x + 1}{3 \left(-\frac{\pi}{2} \cos\left(-\frac{\pi}{2} x\right)\right) + 6x}
\]

Substituting \( x = 1 \):

\[
\frac{3(1)^2 + 8(1) + 1}{3 \left(-\frac{\pi}{2} \cos\left(-\frac{\pi}{2}(1)\right)\right) + 6(1)} = \frac{3 + 8 + 1}{3 \cdot 0 + 6} = \frac{12}{6} = 2
\]

Thus, \( \lim_{x \to 1} f(x) = 2 \).

### Step 3: Conclusion

- The limit \( \lim_{x \to 1} f(x) \) exists and is equal to 2.
- However, \( f(1) \) is not defined due to the indeterminate form \( \frac{0}{0} \).

Thus, the function is not continuous at \( x = 1 \) because the limit exists, but the function value \( f(1) \) does not exist. Therefore, the correct answer is **B**: \( \lim_{x \to 1} f(x) \) exists, but \( f(1) \) does not exist.

---

Do you want more details on any part of the solution, or have any questions? 

Here are five related questions to deepen understanding:
1. What are the three conditions required for a function to be continuous at a point?
2. How does L'Hopital's Rule help in resolving indeterminate forms like \( \frac{0}{0} \)?
3. What are the different types of discontinuities a function can have?
4. Can a function be continuous if its limit does not exist at a particular point?
5. How would you evaluate \( f(x) \) if the denominator were not indeterminate at \( x = 1 \)?

**Tip:** When checking for continuity, always verify both the limit and the function value at the given point.

Let f be the function defined as f(x) = (x^3 + 4x^2 + x - 6) / (3sin(-π/2 * x) + 3x^2). Which of the following conditions explains why f is not continuous at x = 1?

Explore the steps to determine why the function f(x) = (x^3 + 4x^2 + x - 6) / (3sin(-π/2 * x) + 3x^2) is not continuous at x = 1. This solution delves into the concepts of limits, continuity, indeterminate forms, and L'Hopital's Rule. Perfect for students in Grades 11-12.

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