Find all point of discountinuity 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: Identify the function and its expression in terms of \( x \)

Given:
\[ f(t) = \frac{1}{t^2 + t - 2} \]
and
\[ t = \frac{1}{x - 1} \]

### Step 2: Substitute \( t \) into the function

Substituting \( t \) into the function gives:
\[ f\left(\frac{1}{x - 1}\right) = \frac{1}{\left(\frac{1}{x - 1}\right)^2 + \left(\frac{1}{x - 1}\right) - 2} \]

### Step 3: Simplify the expression

Calculating \( \left(\frac{1}{x - 1}\right)^2 \):
\[ \left(\frac{1}{x - 1}\right)^2 = \frac{1}{(x - 1)^2} \]
Thus, we have:
\[ f\left(\frac{1}{x - 1}\right) = \frac{1}{\frac{1}{(x - 1)^2} + \frac{1}{x - 1} - 2} \]

Finding a common denominator for the terms in the denominator:
\[ = \frac{1}{\frac{1 + (x - 1) - 2(x - 1)^2}{(x - 1)^2}} \]
This simplifies to:
\[ = \frac{(x - 1)^2}{1 + (x - 1) - 2(x - 1)^2} \]

### Step 4: Further simplify the denominator

The denominator simplifies to:
\[ 1 + x - 1 - 2(x^2 - 2x + 1) = 1 + x - 1 - 2x^2 + 4x - 2 = -2x^2 + 5x - 2 \]
Thus:
\[ f\left(\frac{1}{x - 1}\right) = \frac{(x - 1)^2}{-2x^2 + 5x - 2} \]

### Step 5: Determine points of discontinuity

The function \( f(t) \) will be discontinuous when the denominator is equal to zero:
\[ -2x^2 + 5x - 2 = 0 \]
We can factor or use the quadratic formula to find the roots:
\[ 2x^2 - 5x + 2 = 0 \]

### Step 6: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
Thus, the solutions are:
\[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{2}{4} = \frac{1}{2} \]

### Step 7: Check for additional points of discontinuity

Since \( t = \frac{1}{x - 1} \), we also need to check when \( x - 1 = 0 \):
\[ x = 1 \]

### Final Points of Discontinuity

The points of discontinuity are:
\[ x = \frac{1}{2}, \quad x = 2, \quad \text{and} \quad x = 1 \]

### Conclusion

Thus, the function \( f(t) \) is discontinuous at \( x = \frac{1}{2}, x = 1, \) and \( x = 2 \).

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