To determine the points of discontinuity in the function $f(x) = \frac{x - 2}{x^2 - 3x + 2}$, we can use the definition of continuity, which states that a function is continuous at a point $x = a$ if the following three conditions hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

The first step is to factor the denominator $x^2 - 3x + 2$.

Step 1: Factor the denominator

The quadratic in the denominator can be factored as: $x^2 - 3x + 2 = (x - 1)(x - 2).$

So the function becomes: $f(x) = \frac{x - 2}{(x - 1)(x - 2)}.$

Step 2: Simplify the function

We notice that the numerator $x - 2$ and one factor in the denominator, $x - 2$, are the same. This means we can cancel the common factor, but we must note that this cancellation is only valid for $x \neq 2$. So the simplified function is: $f(x) = \frac{1}{x - 1}, \quad \text{for} \quad x \neq 2.$

Step 3: Identify points of discontinuity

Now we analyze the points where the function is not continuous:

• At $x = 2$: The original function is undefined because the denominator becomes zero when $x = 2$. Even after simplification, the function still has no definition at $x = 2$. Thus, $x = 2$ is a removable discontinuity, because if we redefine the function at this point, continuity could be restored.

• At $x = 1$: The simplified function $f(x) = \frac{1}{x - 1}$ is undefined at $x = 1$ because the denominator becomes zero. This creates a non-removable (infinite) discontinuity at $x = 1$, as the limit does not exist at this point.

Conclusion

The function has two discontinuities:

1. A removable discontinuity at $x = 2$.
2. A non-removable discontinuity at $x = 1$.

Would you like further details or have any questions?

Here are some related questions to further explore this topic:

1. What is the difference between removable and non-removable discontinuities?
2. How can we redefine the function to remove the discontinuity at $x = 2$?
3. How do you determine the limits at points of discontinuity?
4. Can a function have both types of discontinuities at the same point?
5. How would the graph of this function look around $x = 1$ and $x = 2$?

Tip: To handle rational functions like this one, always check for factors that cancel out in the numerator and denominator, as they often indicate removable discontinuities.