If `f(x)` is defined by: `f(x) = {{:((|x^(2)-x|)/(x^(2)-x), (x ne 0,1)),(1, x=0),(-1, x=1):}` then f(x) is continuous for all

Options

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

Answer

To determine the continuity of the function \( f(x) \) defined as:

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

we need to check the continuity at the points \( x = 0 \) and \( x = 1 \), as well as for other values of \( x \).

### Step 1: Check continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to evaluate:

1. \( f(0) \)
2. \( \lim_{x \to 0^-} f(x) \)
3. \( \lim_{x \to 0^+} f(x) \)

**Calculating \( f(0) \):**
\[ f(0) = 1 \]

**Calculating \( \lim_{x \to 0^-} f(x) \):**

For \( x < 0 \):
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{-(x^2 - x)}{x^2 - x} = 1 \]
Thus,
\[ \lim_{x \to 0^-} f(x) = 1 \]

**Calculating \( \lim_{x \to 0^+} f(x) \):**

For \( x > 0 \):
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{x^2 - x}{x^2 - x} = 1 \]
Thus,
\[ \lim_{x \to 0^+} f(x) = 1 \]

**Conclusion for \( x = 0 \):**
Since
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1, \]
the function is continuous at \( x = 0 \).

### Step 2: Check continuity at \( x = 1 \)

To check continuity at \( x = 1 \), we need to evaluate:

1. \( f(1) \)
2. \( \lim_{x \to 1^-} f(x) \)
3. \( \lim_{x \to 1^+} f(x) \)

**Calculating \( f(1) \):**
\[ f(1) = -1 \]

**Calculating \( \lim_{x \to 1^-} f(x) \):**

For \( x < 1 \):
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{x^2 - x}{x^2 - x} = 1 \]
Thus,
\[ \lim_{x \to 1^-} f(x) = 1 \]

**Calculating \( \lim_{x \to 1^+} f(x) \):**

For \( x > 1 \):
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{x^2 - x}{x^2 - x} = 1 \]
Thus,
\[ \lim_{x \to 1^+} f(x) = 1 \]

**Conclusion for \( x = 1 \):**
Since
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 1 \quad \text{and} \quad f(1) = -1, \]
the function is not continuous at \( x = 1 \).

### Final Conclusion

The function \( f(x) \) is continuous at \( x = 0 \) but not continuous at \( x = 1 \). Therefore, \( f(x) \) is continuous for all \( x \) except at \( x = 1 \).