Find the points of discontinuity of `y = (1)/(u^(2) + u -2)`, where `u = (1)/(x -1)`

Answer

To find the points of discontinuity of the function \( y = \frac{1}{u^2 + u - 2} \) where \( u = \frac{1}{x - 1} \), we will follow these steps:

### Step 1: Substitute \( u \) into the function
Given \( u = \frac{1}{x - 1} \), we can substitute this into the function:
\[ y = \frac{1}{\left(\frac{1}{x - 1}\right)^2 + \left(\frac{1}{x - 1}\right) - 2} \]

### Step 2: Simplify the expression
We first need to simplify the denominator:
\[ u^2 = \left(\frac{1}{x - 1}\right)^2 = \frac{1}{(x - 1)^2} \]
Thus, we can write:
\[ y = \frac{1}{\frac{1}{(x - 1)^2} + \frac{1}{x - 1} - 2} \]
Now, we will find a common denominator for the terms in the denominator:
\[ y = \frac{1}{\frac{1 + (x - 1) - 2(x - 1)^2}{(x - 1)^2}} \]
This simplifies to:
\[ y = \frac{(x - 1)^2}{1 + (x - 1) - 2(x - 1)^2} \]

### Step 3: Further simplify the denominator
Now, let's simplify the expression in the denominator:
\[ 1 + (x - 1) - 2(x - 1)^2 = 1 + x - 1 - 2(x^2 - 2x + 1) = x - 2x^2 + 4x - 2 = -2x^2 + 5x - 2 \]
Thus, we have:
\[ y = \frac{(x - 1)^2}{-2x^2 + 5x - 2} \]

### Step 4: Find points of discontinuity
The function \( y \) will be discontinuous where the denominator is zero or where \( u \) is undefined.

1. **Finding where \( u \) is undefined**:
\[ u = \frac{1}{x - 1} \quad \text{is undefined when } x - 1 = 0 \Rightarrow x = 1 \]

2. **Finding where the denominator is zero**:
We need to solve:
\[ -2x^2 + 5x - 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot (-2) \cdot (-2)}}{2 \cdot (-2)} \]
\[ = \frac{5 \pm \sqrt{25 - 16}}{-4} = \frac{5 \pm 3}{-4} \]
This gives us two solutions:
\[ x = \frac{8}{-4} = -2 \quad \text{and} \quad x = \frac{2}{-4} = -\frac{1}{2} \]

### Final Points of Discontinuity
The points of discontinuity are:
- \( x = 1 \)
- \( x = -2 \)
- \( x = -\frac{1}{2} \)

### Summary of Steps
1. Substitute \( u \) into the function.
2. Simplify the expression to find the denominator.
3. Identify where the denominator is zero and where \( u \) is undefined.
4. Solve for the points of discontinuity.