Step 1: Analyze $f(x)$ and Find Vertical Asymptotes

Given: f(x) = \frac{\frac{x^2 - 2x + 1}{x^2 - 3x + 2}}

Factorize the Numerator and Denominator

1. Numerator $x^2 - 2x + 1$:

   • Factor as $(x - 1)^2$.

2. Denominator $x^2 - 3x + 2$:

   • Factor as $(x - 1)(x - 2)$.

So, we can rewrite $f(x)$ as: $f(x) = \frac{(x - 1)^2}{(x - 1)(x - 2)}$

Simplify the Expression

Cancelling $(x - 1)$ from the numerator and denominator gives: $f(x) = \frac{x - 1}{x - 2}, \quad x \neq 1$

Identify Vertical Asymptotes

The function $f(x)$ will be undefined where the denominator is zero: $x - 2 = 0 \Rightarrow x = 2$

Thus, $x = 2$ is a vertical asymptote.

Limit Notation for the Vertical Asymptote

To find the limit as $x$ approaches the vertical asymptote from the left and right: $\lim_{x \to 2^-} f(x) \quad \text{and} \quad \lim_{x \to 2^+} f(x)$

Since $f(x) = \frac{x - 1}{x - 2}$:

1. As $x \to 2^-$, $f(x)$ approaches $-\infty$.
2. As $x \to 2^+$, $f(x)$ approaches $+\infty$.

Thus, the vertical asymptote can be expressed in limit notation as: $\lim_{x \to 2^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 2^+} f(x) = +\infty$

Step 2: Determine Where $s(x)$ is Undefined

Given: $s(x) = 2f(x - 1) + 2$

Substitute $f(x - 1)$ into $f(x)$ by replacing $x$ with $x - 1$: $f(x - 1) = \frac{(x - 1) - 1}{(x - 1) - 2} = \frac{x - 2}{x - 3}$

Then: $s(x) = 2 \cdot \frac{x - 2}{x - 3} + 2 = \frac{2(x - 2)}{x - 3} + 2$

Find Points Where $s(x)$ is Undefined

The function $s(x)$ will be undefined where the denominator of $\frac{x - 2}{x - 3}$ is zero: $x - 3 = 0 \Rightarrow x = 3$

Therefore, $s(x)$ is undefined at $x = 3$.

Summary

1. Vertical Asymptote of $f(x)$: $x = 2$.
2. Limit Notation for the Asymptote: $\lim_{x \to 2^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 2^+} f(x) = +\infty$
3. Undefined Values of $s(x)$: $x = 3$.

Would you like further clarification on any part of this solution?

Further Related Questions

1. How do you find horizontal asymptotes for rational functions?
2. What happens to $f(x)$ as $x \to \infty$ or $x \to -\infty$?
3. Can we identify removable discontinuities in rational functions?
4. How do we simplify complex rational functions with higher-degree polynomials?
5. How would the result change if the numerator or denominator had additional factors?

Tip

Always check for common factors between the numerator and denominator in rational expressions to identify potential simplifications and removable discontinuities.