`f(x)=x^(3)+3x^(2)-33x-33"for "x gt 0` and g be its inverse such that kg'(2)=1, then the value of k is

Options

A. `-36`
B. `42`
C. 12
D. none of these

Explanation

To solve the problem step by step, we start with the function \( f(x) = x^3 + 3x^2 - 33x - 33 \) for \( x > 0 \) and we need to find the value of \( k \) such that \( k \cdot g'(2) = 1 \), where \( g \) is the inverse of \( f \).

### Step 1: Understand the relationship between \( f \) and \( g \)
Since \( g \) is the inverse of \( f \), we have:
\[ g(f(x)) = x \]
Differentiating both sides with respect to \( x \):
\[ g'(f(x)) \cdot f'(x) = 1 \]
This implies:
\[ g'(f(x)) = \frac{1}{f'(x)} \]

### Step 2: Find \( f'(x) \)
First, we calculate the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 33x - 33) = 3x^2 + 6x - 33 \]

### Step 3: Set up the equation for \( g'(2) \)
From the relationship established, we can substitute \( x \) such that \( f(x) = 2 \):
\[ g'(2) = \frac{1}{f'(x)} \quad \text{where } f(x) = 2 \]

### Step 4: Solve for \( x \) such that \( f(x) = 2 \)
We need to solve the equation:
\[ x^3 + 3x^2 - 33x - 33 = 2 \]
This simplifies to:
\[ x^3 + 3x^2 - 33x - 35 = 0 \]

### Step 5: Factor the polynomial
To factor the polynomial \( x^3 + 3x^2 - 33x - 35 \), we can use the Rational Root Theorem or synthetic division. Testing \( x = -1 \):
\[ (-1)^3 + 3(-1)^2 - 33(-1) - 35 = -1 + 3 + 33 - 35 = 0 \]
So, \( x + 1 \) is a factor. Now we perform synthetic division:
\[ x^3 + 3x^2 - 33x - 35 = (x + 1)(x^2 + 2x - 35) \]
Now we factor \( x^2 + 2x - 35 \):
\[ x^2 + 2x - 35 = (x + 7)(x - 5) \]
Thus, we have:
\[ x^3 + 3x^2 - 33x - 35 = (x + 1)(x + 7)(x - 5) \]

### Step 6: Identify valid solutions
The roots of the equation are \( x = -1, -7, 5 \). Since \( x > 0 \), the only valid solution is \( x = 5 \).

### Step 7: Calculate \( f'(5) \)
Now we calculate \( f'(5) \):
\[ f'(5) = 3(5^2) + 6(5) - 33 = 3(25) + 30 - 33 = 75 + 30 - 33 = 72 \]

### Step 8: Relate \( g'(2) \) to \( k \)
From the earlier relationship, we have:
\[ g'(2) = \frac{1}{f'(5)} = \frac{1}{72} \]
Given \( k \cdot g'(2) = 1 \):
\[ k \cdot \frac{1}{72} = 1 \implies k = 72 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{72} \]