Given the piecewise function [tex]\( f(x) \)[/tex]:

[tex]\[ f(x) = \left\{
\begin{array}{ll}
\frac{x^2 - x}{x - 1} & \text{if } x \neq 1 \\
2 & \text{if } x = 1
\end{array}
\right. \][/tex]



Answer :

Alright, let's carefully explore the function [tex]\( f(x) \)[/tex] defined as follows:

[tex]\[ f(x) = \begin{cases} \frac{x^2 - x}{x - 1} & \text{if } x \neq 1, \\ 2 & \text{if } x = 1. \end{cases} \][/tex]

### Step-by-Step Analysis

1. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x \neq 1 \)[/tex]:

Let's simplify the expression [tex]\(\frac{x^2 - x}{x - 1}\)[/tex].

The numerator [tex]\( x^2 - x \)[/tex] can be factored:
[tex]\[ x^2 - x = x(x - 1). \][/tex]

Substituting this back into the fraction:
[tex]\[ \frac{x^2 - x}{x - 1} = \frac{x(x - 1)}{x - 1}. \][/tex]

For [tex]\( x \neq 1 \)[/tex], the [tex]\((x - 1)\)[/tex] terms in the numerator and denominator cancel out:
[tex]\[ \frac{x(x - 1)}{x - 1} = x. \][/tex]

Therefore, for [tex]\( x \neq 1 \)[/tex]:
[tex]\[ f(x) = x. \][/tex]

2. Evaluate [tex]\( f(x) \)[/tex] specifically at [tex]\( x = 1 \)[/tex]:

By definition in the function:
[tex]\[ f(1) = 2. \][/tex]

3. Verify the consistency and overall understanding:

- For [tex]\( x \neq 1 \)[/tex], [tex]\( f(x) = x \)[/tex].
- At [tex]\( x = 1 \)[/tex], [tex]\( f(1) = 2 \)[/tex].

4. Example Value Testing:

Let's test the function for a value other than 1 to verify consistency (e.g., [tex]\( x = 2 \)[/tex]):

Substitute [tex]\( x = 2 \)[/tex] into the simplified form:
[tex]\[ f(2) = 2. \][/tex]

### Conclusion

Combining all the information:

- If [tex]\( x \neq 1 \)[/tex], then [tex]\( f(x) = x \)[/tex]. For [tex]\( x = 2 \)[/tex], the function value is [tex]\( f(2) = 2 \)[/tex].
- At [tex]\( x = 1 \)[/tex], the function specifies directly that [tex]\( f(1) = 2 \)[/tex].

Thus, the results for the given conditions are:
[tex]\[ f(1) = 2 \][/tex]
[tex]\[ f(2) = 2 \][/tex]

These results help us confirm that the function behaves as expected at and around [tex]\( x = 1 \)[/tex].