f(x)=(x^3+4x^2+x-6)/(3sin(-pi/2x)+3x^2 Let f be the function defined above. Which of the following conditions explains why f is not continuous at x=1? A. Neither lim f(2x) as x goes to 1 nor F (1) exists. B. lim f(x) as x goes to 1 exists, but f (1) does not exist. C. Both lim f(x) as x goes to 1 and f (1) exist, but lim f(x) as x goes to 1 does not equal f(1) D. Both limit f(x) as x goes to 1 and f(1) exist, and limit f(x) as x goes to 1 = f(1)

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.

f(x) = (x^3 + 4x^2 + x - 6) / (3sin(-π/2x) + 3x^2). Let f be the function defined above. Which of the following conditions explains why f is not continuous at x=1? A. Neither lim f(2x) as x goes to 1 nor f(1) exists. B. lim f(x) as x goes to 1 exists, but f(1) does not exist. C. Both lim f(x) as x goes to 1 and f(1) exist, but lim f(x) as x goes to 1 does not equal f(1). D. Both lim f(x) as x goes to 1 and f(1) exist, and limit f(x) as x goes to 1 = f(1).

Explore why the function f(x) = (x^3 + 4x^2 + x - 6) / (3sin(-π/2x) + 3x^2) is not continuous at x=1. Learn about limits, continuity, and L'Hopital's Rule, with a step-by-step explanation. This problem is ideal for students in Grades 11-12 or College Calculus.

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