Let f (x) be a function which satisfy the equation `f (xy) = f(x) + f(y)` for all `x gt 0, y gt 0` such that `f '(1) =2.` Let A be the area of the region bounded by the curves `y =f (x), y = |x ^(3) -6x ^(2)+11 x-6| and x=0,` then find value of ` (28)/(17)A.`

Answer

To solve the problem step-by-step, we need to find the function \( f(x) \) that satisfies the given functional equation and then calculate the area \( A \) bounded by the curves. Finally, we will compute \( \frac{28}{17} A \).

### Step 1: Determine the function \( f(x) \)

The functional equation given is:
\[ f(xy) = f(x) + f(y) \quad \text{for all } x > 0, y > 0 \]
This suggests that \( f(x) \) is a logarithmic function. A common form for such functions is:
\[ f(x) = k \ln(x) \]
for some constant \( k \).

### Step 2: Use the derivative condition

We are given that \( f'(1) = 2 \). First, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(k \ln x) = \frac{k}{x} \]
Now, substituting \( x = 1 \):
\[ f'(1) = k \cdot 1 = k \]
Setting this equal to the given condition:
\[ k = 2 \]
Thus, we have:
\[ f(x) = 2 \ln x \]

### Step 3: Analyze the second function

Next, we need to analyze the function:
\[ y = |x^3 - 6x^2 + 11x - 6| \]
Let:
\[ g(x) = x^3 - 6x^2 + 11x - 6 \]
We will find the roots of \( g(x) \) to understand where the function changes sign.

### Step 4: Find the roots of \( g(x) \)

Using the Rational Root Theorem or synthetic division, we find that \( g(1) = 0 \), so \( x - 1 \) is a factor. Dividing \( g(x) \) by \( x - 1 \):
\[ g(x) = (x - 1)(x^2 - 5x + 6) \]
Factoring the quadratic:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
Thus, the complete factorization is:
\[ g(x) = (x - 1)(x - 2)(x - 3) \]
The roots are \( x = 1, 2, 3 \).

### Step 5: Analyze the behavior of \( g(x) \)

The function \( g(x) \) changes sign at these roots. We can evaluate \( g(x) \) at points around these roots to determine where it is positive or negative:
- For \( x < 1 \): \( g(x) > 0 \)
- For \( 1 < x < 2 \): \( g(x) < 0 \)
- For \( 2 < x < 3 \): \( g(x) > 0 \)
- For \( x > 3 \): \( g(x) > 0 \)

Thus, we have:
\[ |g(x)| = \begin{cases} g(x) & \text{if } x < 1 \\ -g(x) & \text{if } 1 \leq x < 2 \\ g(x) & \text{if } 2 < x < 3 \\ g(x) & \text{if } x > 3 \end{cases} \]

### Step 6: Calculate the area \( A \)

The area \( A \) is bounded by \( f(x) \) and \( |g(x)| \) from \( x = 0 \) to \( x = 3 \).

1. **From \( x = 0 \) to \( x = 1 \)**:
\[ A_1 = \int_0^1 (2 \ln x - (x^3 - 6x^2 + 11x - 6)) \, dx \]

2. **From \( x = 1 \) to \( x = 2 \)**:
\[ A_2 = \int_1^2 (-(2 \ln x) + (x^3 - 6x^2 + 11x - 6)) \, dx \]

3. **From \( x = 2 \) to \( x = 3 \)**:
\[ A_3 = \int_2^3 (2 \ln x - (x^3 - 6x^2 + 11x - 6)) \, dx \]

### Step 7: Compute the integrals

Calculating these integrals will give us the total area \( A \).

### Step 8: Calculate \( \frac{28}{17} A \)

Finally, after computing the total area \( A \), we multiply it by \( \frac{28}{17} \) to find the required value.