Let ` f (x) = x^(3)+ 4x ^(2)+ 6x and g (x)` be inverse then the vlaue of `g' (-4):`

Options

A. `-2`
B. 2
C. `1/2`
D. None of these

Answer

To solve the problem, we need to find the value of \( g'(-4) \) where \( g(x) \) is the inverse of the function \( f(x) = x^3 + 4x^2 + 6x \).

### Step-by-step Solution:

1. **Understanding the relationship between \( f \) and \( g \)**:
Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ g(f(x)) = x \]
Taking the derivative of both sides with respect to \( x \):
\[ g'(f(x)) \cdot f'(x) = 1 \]
This implies:
\[ g'(f(x)) = \frac{1}{f'(x)} \]

2. **Finding \( f'(x) \)**:
First, we need to compute the derivative \( f'(x) \):
\[ f(x) = x^3 + 4x^2 + 6x \]
Using the power rule:
\[ f'(x) = 3x^2 + 8x + 6 \]

3. **Finding \( x \) such that \( f(x) = -4 \)**:
We need to find \( x \) such that:
\[ f(x) = -4 \]
This gives us the equation:
\[ x^3 + 4x^2 + 6x + 4 = 0 \]
Rearranging:
\[ x^3 + 4x^2 + 6x + 4 = 0 \]

4. **Finding the roots of the polynomial**:
We can try \( x = -2 \):
\[ (-2)^3 + 4(-2)^2 + 6(-2) + 4 = -8 + 16 - 12 + 4 = 0 \]
Therefore, \( x = -2 \) is a root.

5. **Finding \( f'(-2) \)**:
Now we substitute \( x = -2 \) into \( f'(x) \):
\[ f'(-2) = 3(-2)^2 + 8(-2) + 6 = 3(4) - 16 + 6 = 12 - 16 + 6 = 2 \]

6. **Finding \( g'(-4) \)**:
Now we can find \( g'(-4) \):
\[ g'(-4) = \frac{1}{f'(-2)} = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( g'(-4) \) is:
\[ \boxed{\frac{1}{2}} \]