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

Question

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

Answer 1

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 2

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

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)