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To determine the continuity and differentiability of the function \( f(x) \) at \( x = \frac{1}{2} \), we need to analyze the following conditions:

### Step 1: Checking Continuity at \( x = \frac{1}{2} \)

The function \( f(x) \) is defined as:
\[
f(x) = \begin{cases} 
x, & \text{when } 0 \leq x < \frac{1}{2} \\
1 - x, & \text{when } \frac{1}{2} \leq x < 1 
\end{cases}
\]

For \( f(x) \) to be continuous at \( x = \frac{1}{2} \), the following must hold:
1. The left-hand limit (LHL) as \( x \to \frac{1}{2}^- \) should exist.
2. The right-hand limit (RHL) as \( x \to \frac{1}{2}^+ \) should exist.
3. The left-hand limit and right-hand limit should be equal to the value of the function at \( x = \frac{1}{2} \).

#### Calculate LHL
Using \( f(x) = x \) when \( x \to \frac{1}{2}^- \):
\[
\lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^-} x = \frac{1}{2}
\]

#### Calculate RHL
Using \( f(x) = 1 - x \) when \( x \to \frac{1}{2}^+ \):
\[
\lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}^+} (1 - x) = 1 - \frac{1}{2} = \frac{1}{2}
\]

#### Evaluate \( f\left(\frac{1}{2}\right) \)
Since \( \frac{1}{2} \leq x < 1 \) applies here, use \( f(x) = 1 - x \) to find:
\[
f\left(\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}
\]

Since \(\text{LHL} = \text{RHL} = f\left(\frac{1}{2}\right) = \frac{1}{2}\), the function is continuous at \( x = \frac{1}{2} \).

### Step 2: Checking Differentiability at \( x = \frac{1}{2} \)

For \( f(x) \) to be differentiable at \( x = \frac{1}{2} \), the left-hand derivative (LHD) and right-hand derivative (RHD) at \( x = \frac{1}{2} \) must be equal.

#### Calculate LHD
\[
\text{LHD} = \lim_{h \to 0^-} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h}
\]
Using \( f(x) = x \) as \( h \to 0^- \):
\[
\text{LHD} = \lim_{h \to 0^-} \frac{\left(\frac{1}{2} + h\right) - \frac{1}{2}}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1
\]

#### Calculate RHD
\[
\text{RHD} = \lim_{h \to 0^+} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h}
\]
Using \( f(x) = 1 - x \) as \( h \to 0^+ \):
\[
\text{RHD} = \lim_{h \to 0^+} \frac{\left(1 - \left(\frac{1}{2} + h\right)\right) - \frac{1}{2}}{h} = \lim_{h \to 0^+} \frac{\frac{1}{2} - h - \frac{1}{2}}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1
\]

Since \(\text{LHD} = 1\) and \(\text{RHD} = -1\), the left-hand and right-hand derivatives are not equal. Thus, the function \( f(x) \) is **not differentiable** at \( x = \frac{1}{2} \).

### Conclusion
- **Continuity**: \( f(x) \) is continuous at \( x = \frac{1}{2} \).
- **Differentiability**: \( f(x) \) is not differentiable at \( x = \frac{1}{2} \).

Would you like further details or have any questions?

### Related Questions
1. How does the continuity of a piecewise function differ from its differentiability?
2. What are the geometric interpretations of continuity and differentiability?
3. Can a function be continuous at a point but not differentiable at that point?
4. How would adding a constant to one piece of the function affect continuity?
5. What are some examples of functions that are continuous everywhere but not differentiable everywhere?

### Tip
For piecewise functions, check continuity at boundary points by ensuring limits from both sides match the function's value.

Test the continuity and differentiability of the function f(x) at x = 1/2, where f(x) is defined as follows:

f(x) = { x, when 0 ≤ x < 1/2
       { 1 - x, when 1/2 ≤ x < 1

Explore the continuity and differentiability of a piecewise function at x = 1/2. This problem involves calculating left-hand and right-hand limits, as well as evaluating left-hand and right-hand derivatives to determine if the function is continuous and differentiable. Suitable for Grades 11-12.

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