The function `f(x) = (x^(2)-1)|x^(2) -3x+2| + cos(|x|)` is not differentiable at

Options

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

Answer

To determine where the function \( f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos(|x|) \) is not differentiable, we will analyze the components of the function step by step.

### Step 1: Identify the components of the function
The function can be broken down into two parts:
- \( g(x) = (x^2 - 1)|x^2 - 3x + 2| \)
- \( h(x) = \cos(|x|) \)

### Step 2: Analyze \( h(x) \)
The function \( h(x) = \cos(|x|) \) is differentiable everywhere since the cosine function is smooth and the absolute value function does not introduce any non-differentiability.

### Step 3: Analyze \( g(x) \)
Next, we need to analyze \( g(x) = (x^2 - 1)|x^2 - 3x + 2| \). The absolute value function \( |x^2 - 3x + 2| \) can introduce points of non-differentiability.

### Step 4: Find the breakpoints of \( |x^2 - 3x + 2| \)
To find the breakpoints, we solve the equation \( x^2 - 3x + 2 = 0 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) = 0 \]
This gives us the breakpoints at \( x = 1 \) and \( x = 2 \).

### Step 5: Check differentiability at the breakpoints
We need to check the differentiability of \( g(x) \) at the breakpoints \( x = 1 \) and \( x = 2 \).

#### At \( x = 1 \):
1. **Left-hand limit** as \( x \) approaches 1:
\[ g(1) = (1^2 - 1)|1^2 - 3(1) + 2| = 0 \cdot |0| = 0 \]
The left-hand derivative:
\[ g'(1^-) = \lim_{h \to 0^-} \frac{g(1 + h) - g(1)}{h} \]
Since \( x^2 - 3x + 2 < 0 \) for \( x < 1 \), we have:
\[ g(x) = (x^2 - 1)(-(x^2 - 3x + 2)) = -(x^2 - 1)(x^2 - 3x + 2) \]

2. **Right-hand limit** as \( x \) approaches 1:
\[ g(1) = 0 \]
The right-hand derivative:
\[ g'(1^+) = \lim_{h \to 0^+} \frac{g(1 + h) - g(1)}{h} \]
Since \( x^2 - 3x + 2 \geq 0 \) for \( x \geq 1 \), we have:
\[ g(x) = (x^2 - 1)(x^2 - 3x + 2) \]

3. Since the left-hand and right-hand derivatives exist and are equal, \( g(x) \) is differentiable at \( x = 1 \).

#### At \( x = 2 \):
1. **Left-hand limit** as \( x \) approaches 2:
\[ g(2) = (2^2 - 1)|2^2 - 3(2) + 2| = 3 \cdot |0| = 0 \]
The left-hand derivative:
\[ g'(2^-) = \lim_{h \to 0^-} \frac{g(2 + h) - g(2)}{h} \]
For \( x < 2 \), \( x^2 - 3x + 2 < 0 \):
\[ g(x) = -(x^2 - 1)(x^2 - 3x + 2) \]

2. **Right-hand limit** as \( x \) approaches 2:
\[ g(2) = 0 \]
The right-hand derivative:
\[ g'(2^+) = \lim_{h \to 0^+} \frac{g(2 + h) - g(2)}{h} \]
For \( x > 2 \), \( x^2 - 3x + 2 \geq 0 \):
\[ g(x) = (x^2 - 1)(x^2 - 3x + 2) \]

3. Calculate both derivatives at \( x = 2 \):
- Left-hand derivative:
\[ g'(2^-) = -4 \cdot 0 + 3 \cdot 1 = -4 \]
- Right-hand derivative:
\[ g'(2^+) = 4 \cdot 1 + 3 \cdot 1 = 7 \]

Since the left-hand derivative does not equal the right-hand derivative at \( x = 2 \), \( g(x) \) is not differentiable at \( x = 2 \).

### Conclusion
The function \( f(x) \) is not differentiable at \( x = 2 \).