`f (x) = {{:((|x^(2)- x|)/(x^(2) - x),xne 0"," 1),(1",", x = 0),(-1",", x = 0 ):}` is continuose for all :

Options

A. x
B. x except at x = 0
C. x except at x = 1
D. x except at = and x = 1

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{|x^2 - x|}{x^2 - x}, & x \neq 0 \\ 1, & x = 0 \\ -1, & x = 1 \end{cases} \]

we need to check the continuity at the points where the function changes its definition, specifically at \(x = 0\) and \(x = 1\).

### Step 1: Analyze the function for \(x \neq 0\)

For \(x \neq 0\), we can simplify the expression:

1. **Identify the intervals**:
- The expression \(x^2 - x = x(x - 1)\) is zero at \(x = 0\) and \(x = 1\).
- The sign of \(x^2 - x\) will change at these points.

2. **Determine the intervals**:
- For \(x < 0\): \(x^2 - x > 0\) (since both \(x\) and \(x - 1\) are negative).
- For \(0 < x < 1\): \(x^2 - x < 0\) (since \(x\) is positive and \(x - 1\) is negative).
- For \(x > 1\): \(x^2 - x > 0\) (since both \(x\) and \(x - 1\) are positive).

3. **Define the function based on these intervals**:
- For \(x < 0\): \(f(x) = \frac{x^2 - x}{x^2 - x} = 1\)
- For \(0 < x < 1\): \(f(x) = \frac{-(x^2 - x)}{x^2 - x} = -1\)
- For \(x > 1\): \(f(x) = \frac{x^2 - x}{x^2 - x} = 1\)

Thus, we can rewrite the function as:

\[ f(x) = \begin{cases} 1, & x < 0 \\ -1, & 0 < x < 1 \\ 1, & x > 1 \end{cases} \]

### Step 2: Check continuity at \(x = 0\)

1. **Calculate the left-hand limit** as \(x\) approaches \(0\) from the left:
\[ \lim_{x \to 0^-} f(x) = 1 \]

2. **Calculate the right-hand limit** as \(x\) approaches \(0\) from the right:
\[ \lim_{x \to 0^+} f(x) = -1 \]

3. **Check the value of the function at \(x = 0\)**:
\[ f(0) = 1 \]

Since the left-hand limit (1) does not equal the right-hand limit (-1), \(f(x)\) is **not continuous at \(x = 0\)**.

### Step 3: Check continuity at \(x = 1\)

1. **Calculate the left-hand limit** as \(x\) approaches \(1\) from the left:
\[ \lim_{x \to 1^-} f(x) = -1 \]

2. **Calculate the right-hand limit** as \(x\) approaches \(1\) from the right:
\[ \lim_{x \to 1^+} f(x) = 1 \]

3. **Check the value of the function at \(x = 1\)**:
\[ f(1) = -1 \]

Since the left-hand limit (-1) does not equal the right-hand limit (1), \(f(x)\) is **not continuous at \(x = 1\)**.

### Conclusion

The function \(f(x)\) is continuous for all \(x \in \mathbb{R}\) except at the points \(x = 0\) and \(x = 1\).

### Final Answer

The function \(f(x)\) is continuous for all \(x \in \mathbb{R}\) except at \(x = 0\) and \(x = 1\).

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