The function `f(x) = (4-x^(2))/(4x-x^(3))` is

Options

A. discontinuous at only one point
B. discontinuous at exactly two points
C. discontinuous at exactly three points
D. None of the above

Explanation

To determine the points of discontinuity for the function \( f(x) = \frac{4 - x^2}{4x - x^3} \), we need to analyze the denominator, as a function is discontinuous where the denominator is equal to zero.

### Step-by-Step Solution:

1. **Identify the Denominator**:
The denominator of the function is \( 4x - x^3 \).

2. **Set the Denominator to Zero**:
To find the points of discontinuity, we set the denominator equal to zero:
\[ 4x - x^3 = 0 \]

3. **Factor the Equation**:
We can factor out \( x \) from the equation:
\[ x(4 - x^2) = 0 \]

4. **Solve for \( x \)**:
This gives us two factors to solve:
- \( x = 0 \)
- \( 4 - x^2 = 0 \)

For the second factor:
\[ 4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm 2 \]

5. **List the Points of Discontinuity**:
From our calculations, we find the points of discontinuity:
- \( x = 0 \)
- \( x = 2 \)
- \( x = -2 \)

6. **Conclusion**:
Therefore, the function \( f(x) \) is discontinuous at exactly three points: \( x = 0, x = 2, \) and \( x = -2 \).

### Final Answer:
The function \( f(x) = \frac{4 - x^2}{4x - x^3} \) is discontinuous at exactly three points: \( x = 0, x = 2, \) and \( x = -2 \).

---