Discuss the differentiability of the function
`f(x)={{:(2x-1,whenx,lt1/2),(3-6x,when x, ge1/2):} at x=1/2.`

Answer

To discuss the differentiability of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} 2x - 1 & \text{when } x < \frac{1}{2} \\ 3 - 6x & \text{when } x \geq \frac{1}{2} \end{cases} \]

at \( x = \frac{1}{2} \), we will follow these steps:

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

To check the continuity of \( f(x) \) at \( x = \frac{1}{2} \), we need to find the left-hand limit, right-hand limit, and the value of the function at that point.

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

2. **Right-hand limit** as \( x \) approaches \( \frac{1}{2} \):
\[ \lim_{x \to \frac{1}{2}^+} f(x) = 3 - 6\left(\frac{1}{2}\right) = 3 - 3 = 0 \]

3. **Value of the function** at \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = 3 - 6\left(\frac{1}{2}\right) = 0 \]

Since the left-hand limit, right-hand limit, and the value of the function at \( x = \frac{1}{2} \) are all equal to 0, we conclude that \( f(x) \) is continuous at \( x = \frac{1}{2} \).

### Step 2: Find the Derivative \( f'(x) \)

Next, we find the derivative of \( f(x) \) for both cases:

1. For \( x < \frac{1}{2} \):
\[ f'(x) = 2 \]

2. For \( x \geq \frac{1}{2} \):
\[ f'(x) = -6 \]

### Step 3: Check Differentiability at \( x = \frac{1}{2} \)

Now, we need to check the left-hand derivative and right-hand derivative at \( x = \frac{1}{2} \):

1. **Left-hand derivative** as \( x \) approaches \( \frac{1}{2} \):
\[ \lim_{x \to \frac{1}{2}^-} f'(x) = 2 \]

2. **Right-hand derivative** as \( x \) approaches \( \frac{1}{2} \):
\[ \lim_{x \to \frac{1}{2}^+} f'(x) = -6 \]

Since the left-hand derivative (2) and the right-hand derivative (-6) are not equal, the derivative \( f'(x) \) is discontinuous at \( x = \frac{1}{2} \).

### Conclusion

The function \( f(x) \) is continuous at \( x = \frac{1}{2} \) but not differentiable at that point because the left-hand and right-hand derivatives do not match.

### Final Answer

The function \( f(x) \) is continuous at \( x = \frac{1}{2} \), but it is not differentiable at \( x = \frac{1}{2} \).