if f(x) =`x-[x], in R,` then `f^(')(1/2)` is equal to

Options

A. `3/2`
B. 1
C. 0
D. `-1`

Explanation

To find \( f'(1/2) \) for the function \( f(x) = x - [x] \), where \([x]\) is the greatest integer less than or equal to \(x\), we will compute the left-hand and right-hand derivatives at \(x = 1/2\).

### Step 1: Define the function
The function is given by:
\[ f(x) = x - [x] \]
For \(x \in [0, 1)\), \([x] = 0\). Thus, for \(x \in [0, 1)\):
\[ f(x) = x - 0 = x \]

### Step 2: Compute the left-hand derivative at \(x = 1/2\)
The left-hand derivative is defined as:
\[ f'_{-}(1/2) = \lim_{h \to 0^-} \frac{f(1/2 + h) - f(1/2)}{h} \]
For \(h < 0\) and \(1/2 + h < 1\):
\[ f(1/2 + h) = 1/2 + h \]
Thus,
\[ f'_{-}(1/2) = \lim_{h \to 0^-} \frac{(1/2 + h) - (1/2)}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]

### Step 3: Compute the right-hand derivative at \(x = 1/2\)
The right-hand derivative is defined as:
\[ f'_{+}(1/2) = \lim_{h \to 0^+} \frac{f(1/2 + h) - f(1/2)}{h} \]
For \(h > 0\) and \(1/2 + h < 1\):
\[ f(1/2 + h) = 1/2 + h \]
Thus,
\[ f'_{+}(1/2) = \lim_{h \to 0^+} \frac{(1/2 + h) - (1/2)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

### Step 4: Conclusion
Since both the left-hand and right-hand derivatives at \(x = 1/2\) are equal:
\[ f'(1/2) = f'_{-}(1/2) = f'_{+}(1/2) = 1 \]
Thus, we conclude that:
\[ f'(1/2) = 1 \]