If `f(x) = {|x|-|x-1|}^(2)`, draw the graph of f(x) and discuss its continuity and differentiability of f(x)

Answer

To solve the problem, we need to analyze the function \( f(x) = (|x| - |x-1|)^2 \). We will break it down into intervals based on the critical points where the absolute values change, which are at \( x = 0 \) and \( x = 1 \).

### Step 1: Define the function in intervals

1. **For \( x < 0 \)**:
- \( |x| = -x \)
- \( |x-1| = 1 - x \) (since \( x < 0 \) implies \( x-1 < 0 \))
- Thus, \( f(x) = (-x - (1 - x))^2 = (-x - 1 + x)^2 = (-1)^2 = 1 \).

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

3. **For \( x \geq 1 \)**:
- \( |x| = x \)
- \( |x-1| = x - 1 \) (since \( x \geq 1 \))
- Thus, \( f(x) = (x - (x - 1))^2 = (1)^2 = 1 \).

### Summary of the function in different intervals:
- For \( x < 0 \): \( f(x) = 1 \)
- For \( 0 \leq x < 1 \): \( f(x) = (2x - 1)^2 \)
- For \( x \geq 1 \): \( f(x) = 1 \)

### Step 2: Graph the function

- For \( x < 0 \), the function is constant at \( f(x) = 1 \).
- For \( 0 \leq x < 1 \), the function is a parabola opening upwards, with its vertex at \( x = \frac{1}{2} \) where \( f(\frac{1}{2}) = 0 \).
- For \( x \geq 1 \), the function is again constant at \( f(x) = 1 \).

### Step 3: Continuity of the function

To check continuity at the critical points \( x = 0 \) and \( x = 1 \):

- **At \( x = 0 \)**:
- \( f(0) = (2(0) - 1)^2 = 1 \)
- Limit as \( x \to 0^- \) is \( 1 \) and limit as \( x \to 0^+ \) is \( 1 \).
- Thus, \( f(x) \) is continuous at \( x = 0 \).

- **At \( x = 1 \)**:
- \( f(1) = 1 \)
- Limit as \( x \to 1^- \) is \( (2(1) - 1)^2 = 1 \) and limit as \( x \to 1^+ \) is \( 1 \).
- Thus, \( f(x) \) is continuous at \( x = 1 \).

Since \( f(x) \) is continuous at all points, it is continuous everywhere.

### Step 4: Differentiability of the function

To check differentiability at the critical points \( x = 0 \) and \( x = 1 \):

- **At \( x = 0 \)**:
- Left-hand derivative:
\[ \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(2h - 1)^2 - 1}{h} = \lim_{h \to 0^-} \frac{4h^2 - 4h}{h} = \lim_{h \to 0^-} (4h - 4) = -4 \]
- Right-hand derivative:
\[ \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0 \]
- Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

- **At \( x = 1 \)**:
- Left-hand derivative:
\[ \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(2(1 + h) - 1)^2 - 1}{h} = \lim_{h \to 0^-} \frac{(2 + 2h - 1)^2 - 1}{h} = \lim_{h \to 0^-} \frac{(1 + 2h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{4h^2 + 4h}{h} = 4 \]
- Right-hand derivative:
\[ \lim_{h \to 0^+} \frac{1 - 1}{h} = 0 \]
- Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

- The function \( f(x) \) is continuous everywhere.
- The function \( f(x) \) is not differentiable at \( x = 0 \) and \( x = 1 \).