The function `f(x) = (x^(2) - 1)|x^(2)-6x + 5|+cos|x|` is not differentiable at

Options

A. -1
B. 0
C. 1
D. 5

Answer

To determine where the function \( f(x) = (x^2 - 1)|x^2 - 6x + 5| + \cos|x| \) 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:
1. \( g(x) = (x^2 - 1)|x^2 - 6x + 5| \)
2. \( h(x) = \cos|x| \)

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

### Step 3: Analyze \( g(x) = (x^2 - 1)|x^2 - 6x + 5| \)
To analyze \( g(x) \), we need to find the points where \( |x^2 - 6x + 5| \) is not differentiable.

#### Step 3.1: Factor the quadratic
The expression \( x^2 - 6x + 5 \) can be factored as:
\[ x^2 - 6x + 5 = (x - 1)(x - 5) \]
The absolute value function \( |x^2 - 6x + 5| \) will change its form at the roots of the quadratic, which are \( x = 1 \) and \( x = 5 \).

### Step 4: Determine the behavior of \( g(x) \)
We need to consider the behavior of \( g(x) \) around the points \( x = 1 \) and \( x = 5 \).

#### Step 4.1: Analyze \( g(x) \) for intervals
1. **For \( x < 1 \)**:
- \( x^2 - 6x + 5 > 0 \) (since both roots are greater than \( x \))
- Thus, \( g(x) = (x^2 - 1)(x^2 - 6x + 5) \)

2. **For \( 1 < x < 5 \)**:
- \( x^2 - 6x + 5 < 0 \)
- Thus, \( g(x) = (x^2 - 1)(-(x^2 - 6x + 5)) = -(x^2 - 1)(x^2 - 6x + 5) \)

3. **For \( x > 5 \)**:
- \( x^2 - 6x + 5 > 0 \)
- Thus, \( g(x) = (x^2 - 1)(x^2 - 6x + 5) \)

### Step 5: Check differentiability at critical points
To check differentiability at \( x = 1 \) and \( x = 5 \), we need to compute the left-hand and right-hand derivatives at these points.

#### Step 5.1: At \( x = 1 \)
- **Left-hand derivative**:
\[ g'(x) = \text{Derivative of } (x^2 - 1)(x^2 - 6x + 5) \]
Evaluate as \( x \to 1^- \).

- **Right-hand derivative**:
\[ g'(x) = \text{Derivative of } -(x^2 - 1)(x^2 - 6x + 5) \]
Evaluate as \( x \to 1^+ \).

Since the expressions will yield different values, \( g(x) \) is not differentiable at \( x = 1 \).

#### Step 5.2: At \( x = 5 \)
- **Left-hand derivative**:
\[ g'(x) = \text{Derivative of } -(x^2 - 1)(x^2 - 6x + 5) \]
Evaluate as \( x \to 5^- \).

- **Right-hand derivative**:
\[ g'(x) = \text{Derivative of } (x^2 - 1)(x^2 - 6x + 5) \]
Evaluate as \( x \to 5^+ \).

Again, since the expressions will yield different values, \( g(x) \) is not differentiable at \( x = 5 \).

### Conclusion
The function \( f(x) \) is not differentiable at:
- \( x = 1 \)
- \( x = 5 \)

### Final Answer
The function \( f(x) \) is not differentiable at \( x = 1 \) and \( x = 5 \).