Question

f(x) = (x - 2) / (x^2 - 3x + 2)

For the function above, find and the discontinuities and identify what types of discontinuities they are

Answer 1

the explanation is not written in normal mathematical notation, may you explain in proper mathematical notation?

Answer 2

To analyze the function \( f(x) = \frac{x - 2}{x^2 - 3x + 2} \), we first need to find the discontinuities by examining the denominator, since a rational function is discontinuous wherever its denominator is zero.

1. **Factor the denominator**:

The denominator is \( x^2 - 3x + 2 \).
We can factor it:
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

2. **Set the denominator equal to zero**:
\[
(x - 1)(x - 2) = 0
\]
This gives us two points where the function could be discontinuous:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

Thus, the potential points of discontinuity are \( x = 1 \) and \( x = 2 \).

3. **Determine the types of discontinuities**:
- At **\( x = 2 \)**:
We notice that both the numerator and denominator become zero:
\[
f(x) = \frac{x - 2}{(x - 1)(x - 2)}
\]
As \( x \) approaches 2, the function approaches:
\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x - 2}{(x - 1)(x - 2)} = \lim_{x \to 2} \frac{1}{x - 1} = \frac{1}{1} = 1
\]
However, since \( f(2) \) is not defined (division by zero), this point is a **removable discontinuity**.

- At **\( x = 1 \)**:
We can evaluate what happens as we approach this point:
\[
\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x - 2}{(x - 1)(x - 2)}
\]
This expression approaches:
\[
\lim_{x \to 1} \frac{1 - 2}{(1 - 1)(1 - 2)} = \frac{-1}{0}
\]
Since the denominator approaches zero while the numerator remains negative, the limit approaches negative infinity or positive infinity, indicating an **infinite discontinuity** (or non-removable discontinuity).

### Summary:
- **Discontinuities**: \( x = 1 \) (infinite discontinuity), \( x = 2 \) (removable discontinuity).

Answer 3

Certainly! Let's analyze the function \( f(x) = \frac{x - 2}{x^2 - 3x + 2} \) and find its discontinuities using proper mathematical notation.

### Step 1: Factor the Denominator

The denominator can be factored as follows:

\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

### Step 2: Identify Potential Discontinuities

We need to determine where the denominator is equal to zero:

\[
(x - 1)(x - 2) = 0
\]

Setting each factor equal to zero gives us:

\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

Thus, the potential points of discontinuity are \( x = 1 \) and \( x = 2 \).

### Step 3: Determine the Types of Discontinuities

#### At \( x = 2 \):

Evaluate the behavior of \( f(x) \) as \( x \) approaches 2. Both numerator and denominator approach zero:

\[
f(x) = \frac{x - 2}{(x - 1)(x - 2)}
\]

Computing the limit as \( x \) approaches 2:

\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x - 2}{(x - 1)(x - 2)} = \lim_{x \to 2} \frac{1}{x - 1}
\]

Evaluating the limit results in:

\[
\lim_{x \to 2} \frac{1}{x - 1} = \frac{1}{2 - 1} = 1
\]

Since \( f(2) \) is not defined, this indicates a **removable discontinuity** at \( x = 2 \).

#### At \( x = 1 \):

Next, evaluate the limit as \( x \) approaches 1:

\[
\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x - 2}{(x - 1)(x - 2)}
\]

As \( x \) approaches 1, the denominator approaches zero:

\[
= \lim_{x \to 1} \frac{1 - 2}{(1 - 1)(1 - 2)} = \frac{-1}{0}
\]

Since the limit does not approach a finite value (it approaches either positive infinity or negative infinity), this indicates an **infinite discontinuity** at \( x = 1 \).

### Conclusion

The function \( f(x) \) has:
- **Removable Discontinuity** at \( x = 2 \)
- **Infinite Discontinuity** at \( x = 1 \)

Thus, we summarize the discontinuities as follows:

1. \( x = 1 \) (infinite discontinuity)
2. \( x = 2 \) (removable discontinuity)