Find all point of discontinuity of the function `f(t)=1/(t^2+t-2),` where `t=1/(x-1)`

Explanation

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

### Step 1: Substitute \( t \) in the function
Given \( t = \frac{1}{x - 1} \), we substitute this into the function:

\[ f\left(\frac{1}{x - 1}\right) = \frac{1}{\left(\frac{1}{x - 1}\right)^2 + \left(\frac{1}{x - 1}\right) - 2} \]

### Step 2: Simplify the expression
We need to simplify the denominator:

\[ \left(\frac{1}{x - 1}\right)^2 = \frac{1}{(x - 1)^2} \]

Thus, the denominator becomes:

\[ \frac{1}{(x - 1)^2} + \frac{1}{x - 1} - 2 \]

To combine these fractions, we take \( (x - 1)^2 \) as the common denominator:

\[ \frac{1 + (x - 1) - 2(x - 1)^2}{(x - 1)^2} \]

### Step 3: Expand and simplify the numerator
Now we expand the numerator:

\[ 1 + (x - 1) - 2(x^2 - 2x + 1) = 1 + x - 1 - 2x^2 + 4x - 2 = -2x^2 + 5x - 2 \]

So, we have:

\[ f\left(\frac{1}{x - 1}\right) = \frac{1}{\frac{-2x^2 + 5x - 2}{(x - 1)^2}} = \frac{(x - 1)^2}{-2x^2 + 5x - 2} \]

### Step 4: Identify points of discontinuity
The function \( f(t) \) will be discontinuous when the denominator is zero:

\[ -2x^2 + 5x - 2 = 0 \]

### Step 5: Solve the quadratic equation
To solve the quadratic equation, we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = -2, b = 5, c = -2 \):

\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot (-2) \cdot (-2)}}{2 \cdot (-2)} \]

Calculating the discriminant:

\[ b^2 - 4ac = 25 - 16 = 9 \]

Now substituting back:

\[ x = \frac{-5 \pm 3}{-4} \]

Calculating the two possible values:

1. \( x = \frac{-5 + 3}{-4} = \frac{-2}{-4} = \frac{1}{2} \)
2. \( x = \frac{-5 - 3}{-4} = \frac{-8}{-4} = 2 \)

### Conclusion
The points of discontinuity are:

\[ x = \frac{1}{2} \quad \text{and} \quad x = 2 \]