The function f(x) is defined as:
`f(x) =1/3 -x, x ,lt 1/3`
`=(1/3-x)^(2), x ge 1/3`
then in.the itnerval (0,1), the mean value, theorem is not true because

Options

A. f(x) is not continuous
B. f(x) is not differentiable
C. `f(0) ne f(1)`
D. none

Answer

To determine why the Mean Value Theorem (MVT) is not applicable for the given function \( f(x) \) in the interval \( (0, 1) \), we need to check two main conditions: continuity and differentiability of the function on the specified interval.

### Step 1: Define the function
The function \( f(x) \) is given as:
\[ f(x) = \begin{cases} \frac{1}{3} - x & \text{if } x < \frac{1}{3} \\ \left(\frac{1}{3} - x\right)^2 & \text{if } x \geq \frac{1}{3} \end{cases} \]

### Step 2: Check continuity
To check continuity at \( x = \frac{1}{3} \), we need to ensure that the left-hand limit and right-hand limit at this point are equal to the function value at \( x = \frac{1}{3} \).

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

2. **Right-hand limit** as \( x \) approaches \( \frac{1}{3} \):
\[ \lim_{x \to \frac{1}{3}^+} f(x) = \left(\frac{1}{3} - \frac{1}{3}\right)^2 = 0 \]

3. **Function value** at \( x = \frac{1}{3} \):
\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3} - \frac{1}{3}\right)^2 = 0 \]

Since the left-hand limit, right-hand limit, and the function value at \( x = \frac{1}{3} \) are all equal to 0, \( f(x) \) is continuous at \( x = \frac{1}{3} \).

### Step 3: Check differentiability
Next, we need to check if \( f(x) \) is differentiable at \( x = \frac{1}{3} \).

1. **Left-hand derivative**:
\[ f'(x) = \frac{d}{dx}\left(\frac{1}{3} - x\right) = -1 \quad \text{for } x < \frac{1}{3} \]
Thus,
\[ \lim_{h \to 0^-} \frac{f\left(\frac{1}{3} + h\right) - f\left(\frac{1}{3}\right)}{h} = -1 \]

2. **Right-hand derivative**:
\[ f'(x) = \frac{d}{dx}\left(\left(\frac{1}{3} - x\right)^2\right) = -2\left(\frac{1}{3} - x\right) \quad \text{for } x > \frac{1}{3} \]
Thus,
\[ \lim_{h \to 0^+} \frac{f\left(\frac{1}{3} + h\right) - f\left(\frac{1}{3}\right)}{h} = -2\left(\frac{1}{3} - \frac{1}{3}\right) = 0 \]

Since the left-hand derivative (-1) does not equal the right-hand derivative (0), \( f(x) \) is not differentiable at \( x = \frac{1}{3} \).

### Conclusion
The Mean Value Theorem requires that the function be continuous on the closed interval and differentiable on the open interval. Although \( f(x) \) is continuous on \( [0, 1] \), it is not differentiable at \( x = \frac{1}{3} \). Therefore, the Mean Value Theorem does not hold for \( f(x) \) in the interval \( (0, 1) \).

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