The function `f(x) = {{:(|x-3|, x ge 1),(x^(2)//4-3x//2 + 13//4, x lt 1):}` is

Options

A. continous at x=1
B. continous at x=3
C. differentiable at x=1
D. differentiable at x=3

Answer

To determine the continuity and differentiability of the function

\[ f(x) = \begin{cases} |x - 3| & \text{if } x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & \text{if } x < 1 \end{cases} \]

we will analyze the function at the points \(x = 1\) and \(x = 3\).

### Step 1: Check Continuity at \(x = 1\)

1. **Left-hand limit as \(x\) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1} \left(\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}\right) \]
Substituting \(x = 1\):
\[ = \frac{1^2}{4} - \frac{3 \cdot 1}{2} + \frac{13}{4} = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} \]
Finding a common denominator (4):
\[ = \frac{1 - 6 + 13}{4} = \frac{8}{4} = 2 \]

2. **Right-hand limit as \(x\) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1} |x - 3| = |1 - 3| = 2 \]

3. **Value of the function at \(x = 1\)**:
\[ f(1) = |1 - 3| = 2 \]

Since the left-hand limit, right-hand limit, and the value of the function at \(x = 1\) are all equal to 2, we conclude that \(f(x)\) is continuous at \(x = 1\).

### Step 2: Check Continuity at \(x = 3\)

1. **Left-hand limit as \(x\) approaches 3**:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3} |x - 3| = |3 - 3| = 0 \]

2. **Right-hand limit as \(x\) approaches 3**:
\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3} |x - 3| = |3 - 3| = 0 \]

3. **Value of the function at \(x = 3\)**:
\[ f(3) = |3 - 3| = 0 \]

Since the left-hand limit, right-hand limit, and the value of the function at \(x = 3\) are all equal to 0, we conclude that \(f(x)\) is continuous at \(x = 3\).

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

1. **Left-hand derivative at \(x = 1\)**:
\[ f'(1^-) = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{\left(\frac{(1 + h)^2}{4} - \frac{3(1 + h)}{2} + \frac{13}{4}\right) - 2}{h} \]
Calculating the derivative:
\[ = \lim_{h \to 0^-} \frac{\frac{1 + 2h + h^2}{4} - \frac{3 + 3h}{2} + \frac{13}{4} - 2}{h} \]
This simplifies to \(-1\).

2. **Right-hand derivative at \(x = 1\)**:
\[ f'(1^+) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{|1 + h - 3| - 2}{h} = \lim_{h \to 0^+} \frac{-(2 + h)}{h} = -1 \]

Since both the left-hand and right-hand derivatives at \(x = 1\) are equal, \(f(x)\) is differentiable at \(x = 1\).

### Step 4: Check Differentiability at \(x = 3\)

1. **Left-hand derivative at \(x = 3\)**:
\[ f'(3^-) = \lim_{h \to 0^-} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0^-} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]

2. **Right-hand derivative at \(x = 3\)**:
\[ f'(3^+) = \lim_{h \to 0^+} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0^+} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

Since the left-hand and right-hand derivatives at \(x = 3\) are not equal (the left-hand derivative is 1 and the right-hand derivative is 1), we conclude that \(f(x)\) is not differentiable at \(x = 3\).

### Conclusion

1. \(f(x)\) is continuous at \(x = 1\).
2. \(f(x)\) is continuous at \(x = 3\).
3. \(f(x)\) is differentiable at \(x = 1\).
4. \(f(x)\) is not differentiable at \(x = 3\).