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

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 of the function since a function is discontinuous where its denominator is equal to zero.

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = \frac{4 - x^2}{4x - x^3} \]

2. **Factor the denominator**:
We can factor out \( x \) from the denominator:
\[ 4x - x^3 = x(4 - x^2) \]
Thus, we rewrite the function as:
\[ f(x) = \frac{4 - x^2}{x(4 - x^2)} \]

3. **Set the denominator equal to zero**:
For the function to be discontinuous, the denominator must be zero:
\[ x(4 - x^2) = 0 \]

4. **Solve for \( x \)**:
This equation can be solved by setting each factor to zero:
- First factor: \( x = 0 \)
- Second factor: \( 4 - x^2 = 0 \)

Solving \( 4 - x^2 = 0 \):
\[ x^2 = 4 \implies x = \pm 2 \]

5. **List the points of discontinuity**:
The points where the function is discontinuous are:
\[ x = 0, \quad x = 2, \quad x = -2 \]

### Final Answer:
The function \( f(x) \) is discontinuous at \( x = 0, 2, -2 \).

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