To determine why the function $f(x) = \frac{x^3 + 4x^2 + x - 6}{3 \sin(-\frac{\pi}{2}x) + 3x^2}$ is not continuous at $x = 1$, we need to examine the limit of $f(x)$ as $x$ approaches 1 and whether $f(1)$ exists.

A function $f(x)$ is continuous at $x = a$ if:

1. $f(a)$ is defined.
2. The limit $\lim_{{x \to a}} f(x)$ exists.
3. $\lim_{{x \to a}} f(x) = f(a)$.

Let's analyze each part:

Step 1: Check if $f(1)$ is defined

To find $f(1)$, we substitute $x = 1$ into the function:

$f(1) = \frac{1^3 + 4 \cdot 1^2 + 1 - 6}{3 \sin(-\frac{\pi}{2} \cdot 1) + 3 \cdot 1^2}$

Simplify the numerator:

$1^3 + 4 \cdot 1^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0$

Now, simplify the denominator:

$3 \sin(-\frac{\pi}{2}) + 3 \cdot 1^2 = 3 \cdot (-1) + 3 = -3 + 3 = 0$

Both the numerator and the denominator are 0 at $x = 1$, which implies an indeterminate form $\frac{0}{0}$. Hence, $f(1)$ is not defined.

Step 2: Determine if $\lim_{{x \to 1}} f(x)$ exists

To check if the limit exists, we need to analyze the limit of the function as $x$ approaches 1. Since both the numerator and denominator go to 0, we should apply L'Hopital's Rule, which is applicable for limits of the form $\frac{0}{0}$.

Let's compute the derivatives of the numerator and denominator:

• Numerator: $x^3 + 4x^2 + x - 6$

$\frac{d}{dx}(x^3 + 4x^2 + x - 6) = 3x^2 + 8x + 1$

• Denominator: $3 \sin(-\frac{\pi}{2}x) + 3x^2$

$\frac{d}{dx}(3 \sin(-\frac{\pi}{2}x) + 3x^2) = 3 \cdot \cos(-\frac{\pi}{2}x) \cdot (-\frac{\pi}{2}) + 6x = -\frac{3\pi}{2} \cos(-\frac{\pi}{2}x) + 6x$

Now, evaluate the limit as $x$ approaches 1:

$\lim_{{x \to 1}} \frac{3x^2 + 8x + 1}{-\frac{3\pi}{2} \cos(-\frac{\pi}{2} \cdot 1) + 6x}$

Substitute $x = 1$:

$\text{Numerator: } 3 \cdot 1^2 + 8 \cdot 1 + 1 = 3 + 8 + 1 = 12$

$\text{Denominator: } -\frac{3\pi}{2} \cdot \cos(-\frac{\pi}{2}) + 6 \cdot 1 = -\frac{3\pi}{2} \cdot 0 + 6 = 6$

So, the limit is:

$\lim_{{x \to 1}} f(x) = \frac{12}{6} = 2$

Conclusion

• $f(1)$ does not exist (since it results in an indeterminate form).
• $\lim_{{x \to 1}} f(x) = 2$ does exist.

Therefore, the correct answer is:

B. $\lim_{{x \to 1}} f(x)$ exists, but $f(1)$ does not exist.

Would you like more details or have any questions? Here are some related topics you might be interested in:

1. How to apply L'Hopital's Rule in different scenarios.
2. Understanding indeterminate forms in calculus.
3. Conditions for continuity of a function.
4. Exploring trigonometric limits and derivatives.
5. Techniques for computing complex limits.

Tip: When dealing with limits and continuity, always check if the function is defined at the point of interest before proceeding with limit calculations.