Let `f(x) =x-|x-x^(2)|, x in [-1,1]`.Then the number of points at which f(x) is discontinuous is:

Options

A. 0
B. 1
C. 2
D. none

Answer

To determine the number of points at which the function \( f(x) = x - |x - x^2| \) is discontinuous on the interval \([-1, 1]\), we will analyze the function step by step.

### Step 1: Analyze the expression inside the absolute value
The expression inside the absolute value is \( x - x^2 \). We can rewrite it as:
\[ x - x^2 = x(1 - x) \]
This expression is a quadratic function that opens downwards and has roots at \( x = 0 \) and \( x = 1 \).

### Step 2: Determine the critical points
The critical points where the expression changes sign are \( x = 0 \) and \( x = 1 \). We will check the sign of \( x - x^2 \) in the intervals:
1. \( x < 0 \)
2. \( 0 \leq x < 1 \)
3. \( x \geq 1 \)

### Step 3: Evaluate the function in different intervals
- **For \( x < 0 \)**:
Here, \( x - x^2 < 0 \). Therefore, \( |x - x^2| = -(x - x^2) = -x + x^2 \).
Thus, \( f(x) = x - (-x + x^2) = x + x - x^2 = 2x - x^2 \).

- **For \( 0 \leq x < 1 \)**:
Here, \( x - x^2 \geq 0 \). Therefore, \( |x - x^2| = x - x^2 \).
Thus, \( f(x) = x - (x - x^2) = x^2 \).

- **For \( x = 1 \)**:
Here, \( f(1) = 1 - |1 - 1^2| = 1 - |1 - 1| = 1 - 0 = 1 \).

### Step 4: Check continuity at critical points
Now we will check the continuity at the critical points \( x = 0 \) and \( x = 1 \).

- **At \( x = 0 \)**:
- Left-hand limit as \( x \to 0^- \):
\[ f(0^-) = 2(0) - (0)^2 = 0 \]
- Right-hand limit as \( x \to 0^+ \):
\[ f(0^+) = (0)^2 = 0 \]
- Value of the function at \( x = 0 \):
\[ f(0) = 0^2 = 0 \]
Since \( f(0^-) = f(0^+) = f(0) = 0 \), \( f(x) \) is continuous at \( x = 0 \).

- **At \( x = 1 \)**:
- Left-hand limit as \( x \to 1^- \):
\[ f(1^-) = (1)^2 = 1 \]
- Right-hand limit as \( x \to 1^+ \):
\[ f(1^+) = 1 - |1 - 1^2| = 1 - 0 = 1 \]
- Value of the function at \( x = 1 \):
\[ f(1) = 1 \]
Since \( f(1^-) = f(1^+) = f(1) = 1 \), \( f(x) \) is continuous at \( x = 1 \).

### Conclusion
Since \( f(x) \) is continuous at both critical points \( x = 0 \) and \( x = 1 \), and there are no other points of discontinuity in the interval \([-1, 1]\), we conclude that the function \( f(x) \) is continuous throughout the interval.

Thus, the number of points at which \( f(x) \) is discontinuous is:
\[ \boxed{0} \]