The number of point of discountinuity of the rational function `f(x)=(x^(2)-3x+2)/(4x-x^(3))` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the number of points of discontinuity of the rational function \( f(x) = \frac{x^2 - 3x + 2}{4x - x^3} \), we will follow these steps:

### Step 1: Factor the numerator and denominator
The numerator is a quadratic expression:
\[ x^2 - 3x + 2 \]
We can factor this as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

The denominator is:
\[ 4x - x^3 = -x^3 + 4x = -x(x^2 - 4) = -x(x - 2)(x + 2) \]

### Step 2: Rewrite the function
Now we can rewrite \( f(x) \):
\[ f(x) = \frac{(x - 1)(x - 2)}{-x(x - 2)(x + 2)} \]

### Step 3: Simplify the function
Notice that \( (x - 2) \) appears in both the numerator and denominator, so we can cancel it out (except at the point where \( x = 2 \)):
\[ f(x) = \frac{x - 1}{-x(x + 2)} \quad \text{for } x \neq 2 \]

### Step 4: Identify points of discontinuity
The points of discontinuity occur where the denominator is zero or where we canceled terms. The denominator \( -x(x + 2) \) is zero when:
1. \( x = 0 \)
2. \( x = -2 \)

Additionally, we have a removable discontinuity at \( x = 2 \) since we canceled \( (x - 2) \).

### Step 5: List the points of discontinuity
Thus, the points of discontinuity are:
1. \( x = 0 \) (infinite discontinuity)
2. \( x = -2 \) (infinite discontinuity)
3. \( x = 2 \) (removable discontinuity)

### Conclusion
Therefore, the total number of points of discontinuity is 3.

### Final Answer
The number of points of discontinuity of the rational function is **3**.

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