Find the number of values of x for which the function `f(x) = sin pix|(x-1)(x-2)(x-3)|` is non -differentiable.

Answer

To determine the number of values of \( x \) for which the function \( f(x) = \sin(\pi x) |(x-1)(x-2)(x-3)| \) is non-differentiable, we will analyze the function step by step.

### Step 1: Identify points where the function may be non-differentiable
The function \( f(x) \) consists of two parts: \( \sin(\pi x) \) and \( |(x-1)(x-2)(x-3)| \). The absolute value function \( |(x-1)(x-2)(x-3)| \) can cause non-differentiability at the points where the expression inside the absolute value is zero, which are \( x = 1, 2, 3 \).

### Step 2: Check the behavior of \( f(x) \) around the critical points
Now we will check the differentiability of \( f(x) \) at these points.

#### At \( x = 1 \):
- For \( x < 1 \): \( (x-1)(x-2)(x-3) < 0 \) (the product is negative)
- For \( x > 1 \): \( (x-1)(x-2)(x-3) \geq 0 \) (the product is non-negative)

Thus, \( f(x) \) changes from \( -\sin(\pi x)(x-1)(x-2)(x-3) \) for \( x < 1 \) to \( \sin(\pi x)(x-1)(x-2)(x-3) \) for \( x > 1 \).

#### At \( x = 2 \):
- For \( x < 2 \): \( (x-1)(x-2)(x-3) \geq 0 \)
- For \( x > 2 \): \( (x-1)(x-2)(x-3) < 0 \)

Thus, \( f(x) \) changes from \( \sin(\pi x)(x-1)(x-2)(x-3) \) for \( x < 2 \) to \( -\sin(\pi x)(x-1)(x-2)(x-3) \) for \( x > 2 \).

#### At \( x = 3 \):
- For \( x < 3 \): \( (x-1)(x-2)(x-3) \geq 0 \)
- For \( x > 3 \): \( (x-1)(x-2)(x-3) < 0 \)

Thus, \( f(x) \) changes from \( \sin(\pi x)(x-1)(x-2)(x-3) \) for \( x < 3 \) to \( -\sin(\pi x)(x-1)(x-2)(x-3) \) for \( x > 3 \).

### Step 3: Check continuity at the critical points
To check if \( f(x) \) is differentiable at these points, we need to ensure that the left-hand and right-hand limits of the derivative exist and are equal.

- **At \( x = 1 \)**:
- \( f(1) = 0 \)
- Left-hand limit and right-hand limit both yield \( 0 \).

- **At \( x = 2 \)**:
- \( f(2) = 0 \)
- Left-hand limit and right-hand limit both yield \( 0 \).

- **At \( x = 3 \)**:
- \( f(3) = 0 \)
- Left-hand limit and right-hand limit both yield \( 0 \).

### Conclusion
Since the limits from both sides at each of the points \( x = 1, 2, 3 \) are equal and the function is continuous at these points, we conclude that:

The function \( f(x) \) is differentiable at \( x = 1, 2, 3 \).

### Final Answer
Thus, the number of values of \( x \) for which the function \( f(x) \) is non-differentiable is **0**.

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