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We are given the function \( f(x) = \frac{x^3 - 3x}{x^3 + 4} \), and we need to determine which statement about \( f \) at \( x = 2 \) and \( x = -1 \) is true.

### Step 1: Analyze discontinuities
#### 1. Check for discontinuities at \( x = 2 \) and \( x = -1 \)
Discontinuities typically arise where the denominator of the function is zero, as this would make the function undefined.

The denominator of \( f(x) \) is:
\[
x^3 + 4
\]
This expression is never zero for any real value of \( x \), because \( x^3 + 4 = 0 \) has no real solutions (the cubic equation has complex roots). Hence, there are no vertical asymptotes or undefined points caused by division by zero.

Therefore, we conclude there is **no vertical asymptote** or discontinuity due to division by zero at either \( x = 2 \) or \( x = -1 \).

#### 2. Check for removable or jump discontinuities
Since there are no points where the function becomes undefined, we next need to check for continuity or other types of discontinuities (e.g., jump discontinuities).

### Step 2: Evaluate continuity at \( x = -1 \) and \( x = 2 \)
We evaluate the function at \( x = -1 \) and \( x = 2 \):

- For \( x = -1 \):
  \[
  f(-1) = \frac{(-1)^3 - 3(-1)}{(-1)^3 + 4} = \frac{-1 + 3}{-1 + 4} = \frac{2}{3}
  \]
  So, \( f(-1) \) exists and is finite, meaning \( f \) is continuous at \( x = -1 \).

- For \( x = 2 \):
  \[
  f(2) = \frac{(2)^3 - 3(2)}{(2)^3 + 4} = \frac{8 - 6}{8 + 4} = \frac{2}{12} = \frac{1}{6}
  \]
  Similarly, \( f(2) \) exists and is finite, meaning \( f \) is continuous at \( x = 2 \).

### Step 3: Eliminate incorrect answers
Given that the function is continuous at both \( x = 2 \) and \( x = -1 \), we can eliminate options that suggest a discontinuity at these points.

- **Option A** says there is a jump discontinuity at \( x = 2 \), which is false, as we've shown continuity at \( x = 2 \).
- **Option B** also says there is a jump discontinuity at \( x = 2 \), which is false.
- **Option C** mentions a vertical asymptote at \( x = 2 \), which is false.
- **Option D** also mentions a vertical asymptote, which is false.

### Conclusion:
None of the given answer options seem to correctly describe the behavior of \( f(x) \) at \( x = 2 \) and \( x = -1 \). The function is continuous at both points and does not exhibit any discontinuities at these values.

Would you like further details or clarification?

Here are five relative questions for expanding this discussion:
1. What are the different types of discontinuities that can occur in a function?
2. How do we determine whether a function has a removable discontinuity?
3. Can there be a vertical asymptote if the denominator of a rational function is never zero?
4. How does continuity affect the overall behavior of a function at a point?
5. Can a function have different types of discontinuities at different points?

**Tip:** To check for continuity at a point, always verify that the function is defined at the point and that the limit from both sides equals the function's value.

Let f(x) = (x^3 - 3x)/(x^3 + 4). Which of the following statements about f at x = 2 and x = -1 is true?

Analyze the function f(x) = (x^3 - 3x)/(x^3 + 4) to determine the nature of discontinuities at x = 2 and x = -1. Learn how to check for vertical asymptotes, jump discontinuities, and removable discontinuities in rational functions. Suitable for high school students.

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