Draw the graph of the function `f(x) = x - |x - x^(2)|, -1 le x le 1` and discuss the continuity or discontinuity of f in the interval `-1 le x le 1`

Answer

To solve the problem of drawing the graph of the function \( f(x) = x - |x - x^2| \) for the interval \( -1 \leq x \leq 1 \) and discussing its continuity, we will follow these steps:

### Step 1: Analyze the expression inside the absolute value
The function involves an absolute value, so we need to analyze the expression \( x - x^2 \):
\[ x - x^2 = x(1 - x) \]
This expression is zero when \( x = 0 \) and \( x = 1 \).

### Step 2: Determine the sign of \( x - x^2 \)
We will check the sign of \( x - x^2 \) in the intervals defined by the roots:
- For \( x < 0 \): Choose \( x = -1 \):
\[ -1 - (-1)^2 = -1 - 1 = -2 \quad (\text{negative}) \]
- For \( 0 < x < 1 \): Choose \( x = 0.5 \):
\[ 0.5 - (0.5)^2 = 0.5 - 0.25 = 0.25 \quad (\text{positive}) \]
- For \( x > 1 \): Choose \( x = 2 \):
\[ 2 - 2^2 = 2 - 4 = -2 \quad (\text{negative}) \]

Thus, we conclude:
- \( x - x^2 < 0 \) for \( x < 0 \) and \( x > 1 \)
- \( x - x^2 \geq 0 \) for \( 0 \leq x \leq 1 \)

### Step 3: Rewrite the function based on the sign
Now we can rewrite \( f(x) \) based on the sign of \( x - x^2 \):
- For \( x < 0 \):
\[ f(x) = x - (-(x - x^2)) = x + x - x^2 = 2x - x^2 \]
- For \( 0 \leq x \leq 1 \):
\[ f(x) = x - (x - x^2) = x - x + x^2 = x^2 \]

### Step 4: Define the function piecewise
Thus, we can define the function \( f(x) \) as:
\[ f(x) = \begin{cases} 2x - x^2 & \text{for } -1 \leq x < 0 \\ x^2 & \text{for } 0 \leq x \leq 1 \end{cases} \]

### Step 5: Evaluate the function at the endpoints and check continuity
1. At \( x = -1 \):
\[ f(-1) = 2(-1) - (-1)^2 = -2 - 1 = -3 \]
2. At \( x = 0 \):
\[ f(0) = 0^2 = 0 \]
3. At \( x = 1 \):
\[ f(1) = 1^2 = 1 \]

### Step 6: Check continuity at \( x = 0 \)
To check continuity at \( x = 0 \):
- The left-hand limit as \( x \) approaches 0 from the left:
\[ \lim_{x \to 0^-} f(x) = 2(0) - (0)^2 = 0 \]
- The right-hand limit as \( x \) approaches 0 from the right:
\[ \lim_{x \to 0^+} f(x) = 0^2 = 0 \]
- Since \( f(0) = 0 \), \( f(x) \) is continuous at \( x = 0 \).

### Step 7: Conclusion
The function \( f(x) \) is continuous in the interval \( -1 \leq x \leq 1 \).