Consider the function `f(x)=(x-2)|x^(2)-3x+2|`, then the incorrect statement is

Options

A. f(x) is continuous at x = 1
B. f(x) is continuous at x = 2
C. f(x) is differentiable at x = 1
D. f(x) is differentiable at x = 2

Answer

To solve the problem, we need to analyze the function \( f(x) = (x-2)|x^2 - 3x + 2| \) and determine the continuity and differentiability of the function at specific points.

### Step 1: Factor the expression inside the absolute value
The expression \( x^2 - 3x + 2 \) can be factored as:
\[ x^2 - 3x + 2 = (x-1)(x-2) \]
Thus, we can rewrite the function as:
\[ f(x) = (x-2)| (x-1)(x-2) | \]

### Step 2: Identify critical points
The critical points where the expression inside the absolute value changes sign are \( x = 1 \) and \( x = 2 \). We will analyze the function at these points.

### Step 3: Check continuity at \( x = 1 \)
To check the continuity at \( x = 1 \), we need to evaluate the left-hand limit and right-hand limit as \( x \) approaches 1.

**Left-hand limit as \( x \to 1^- \):**
For \( x < 1 \), \( |(x-1)(x-2)| = -(x-1)(x-2) \):
\[ f(x) = (x-2)(-(x-1)(x-2)) = -(x-2)^2(x-1) \]
Calculating the limit:
\[ \lim_{x \to 1^-} f(x) = -(1-2)^2(1-1) = 0 \]

**Right-hand limit as \( x \to 1^+ \):**
For \( x > 1 \), \( |(x-1)(x-2)| = (x-1)(x-2) \):
\[ f(x) = (x-2)((x-1)(x-2)) = (x-2)(x-1)(x-2) \]
Calculating the limit:
\[ \lim_{x \to 1^+} f(x) = (1-2)(1-1)(1-2) = 0 \]

Since both limits equal \( 0 \), \( f(x) \) is continuous at \( x = 1 \).

### Step 4: Check continuity at \( x = 2 \)
**Left-hand limit as \( x \to 2^- \):**
For \( x < 2 \), \( |(x-1)(x-2)| = -(x-1)(x-2) \):
\[ f(x) = (x-2)(-(x-1)(x-2)) = -(x-2)^2(x-1) \]
Calculating the limit:
\[ \lim_{x \to 2^-} f(x) = -(2-2)^2(2-1) = 0 \]

**Right-hand limit as \( x \to 2^+ \):**
For \( x > 2 \), \( |(x-1)(x-2)| = (x-1)(x-2) \):
\[ f(x) = (x-2)((x-1)(x-2)) = (x-2)(x-1)(x-2) \]
Calculating the limit:
\[ \lim_{x \to 2^+} f(x) = (2-2)(2-1)(2-2) = 0 \]

Since both limits equal \( 0 \), \( f(x) \) is continuous at \( x = 2 \).

### Step 5: Check differentiability at \( x = 1 \)
To check differentiability at \( x = 1 \), we need to find the left-hand derivative and right-hand derivative.

**Left-hand derivative as \( x \to 1^- \):**
Using the expression from earlier:
\[ f'(x) = \frac{d}{dx}[-(x-2)^2(x-1)] \]
Calculating the derivative and evaluating at \( x = 1 \):
\[ f'(1^-) = -2(1-2)(1-1) + (1-2)^2 = 0 + 1 = 1 \]

**Right-hand derivative as \( x \to 1^+ \):**
Using the expression from earlier:
\[ f'(x) = \frac{d}{dx}[(x-2)(x-1)(x-2)] \]
Calculating the derivative and evaluating at \( x = 1 \):
\[ f'(1^+) = (1-2)(1-1) + (1-2)(1-2) + (1-2)(1-1) = 0 + 1 + 0 = -1 \]

Since \( f'(1^-) \neq f'(1^+) \), \( f(x) \) is not differentiable at \( x = 1 \).

### Step 6: Check differentiability at \( x = 2 \)
Using similar steps as above, we find that both left-hand and right-hand derivatives at \( x = 2 \) equal \( 0 \), thus \( f(x) \) is differentiable at \( x = 2 \).

### Conclusion
The incorrect statement regarding the function \( f(x) \) is that it is differentiable at \( x = 1 \).