If `f (x )= (x-1) ^(4) (x-2) ^(3) (x-3) ^(2)` then the value of `f '(1) +f''(2) +f''(3)` is:

Options

A. 0
B. 1
C. 2
D. 6

Answer

To solve the problem, we need to find the values of \( f'(1) + f''(2) + f''(3) \) for the function \( f(x) = (x-1)^4 (x-2)^3 (x-3)^2 \).

### Step 1: Find \( f'(x) \)

Using the product rule for differentiation, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( (x-1)^4 (x-2)^3 (x-3)^2 \right) \]

Applying the product rule, we differentiate each factor:

\[ f'(x) = (x-1)^4 \cdot \frac{d}{dx}((x-2)^3 (x-3)^2) + (x-2)^3 (x-3)^2 \cdot \frac{d}{dx}((x-1)^4) \]

Calculating \( \frac{d}{dx}((x-2)^3 (x-3)^2) \) using the product rule again:

\[ \frac{d}{dx}((x-2)^3 (x-3)^2) = (x-2)^3 \cdot \frac{d}{dx}((x-3)^2) + (x-3)^2 \cdot \frac{d}{dx}((x-2)^3) \]

Calculating the derivatives:

\[ \frac{d}{dx}((x-3)^2) = 2(x-3) \]
\[ \frac{d}{dx}((x-2)^3) = 3(x-2)^2 \]

Thus,

\[ \frac{d}{dx}((x-2)^3 (x-3)^2) = (x-2)^3 \cdot 2(x-3) + (x-3)^2 \cdot 3(x-2)^2 \]

### Step 2: Evaluate \( f'(1) \)

Now we substitute \( x = 1 \):

\[ f'(1) = (1-1)^4 \cdot \text{(some expression)} + (1-2)^3 (1-3)^2 \cdot 4(1-1)^3 \]

Since \( (1-1)^4 = 0 \) and \( (1-1)^3 = 0 \), we find that:

\[ f'(1) = 0 \]

### Step 3: Find \( f''(x) \)

Next, we need to find \( f''(x) \). We can differentiate \( f'(x) \) again. However, we can also use the fact that \( f(x) \) has roots at \( x = 1, 2, 3 \) to simplify our calculations.

### Step 4: Evaluate \( f''(2) \) and \( f''(3) \)

Since \( f(x) \) has a factor of \( (x-2)^3 \), when we differentiate it twice, \( f''(2) \) will also yield \( 0 \) because the second derivative will still have a factor of \( (x-2) \).

Similarly, \( f''(3) \) will yield \( 0 \) because of the factor \( (x-3)^2 \).

### Step 5: Combine the results

Now we can combine the results:

\[ f'(1) + f''(2) + f''(3) = 0 + 0 + 0 = 0 \]

### Final Answer

Thus, the value of \( f'(1) + f''(2) + f''(3) \) is:

\[ \boxed{0} \]