If `f (x) =3x ^(9) -2x ^(4) +2x ^(3)-3x ^(2) +x+ cosx +5 and g (x) =f ^(-1) (x),` then the value of `g'(6)` equals:

Options

A. 1
B. `1/2`
C. 2
D. 3

Answer

To find the value of \( g'(6) \) where \( g(x) = f^{-1}(x) \) and \( f(x) = 3x^9 - 2x^4 + 2x^3 - 3x^2 + x + \cos x + 5 \), we can follow these steps:

### Step 1: Understand the relationship between \( g \) and \( f \)
Since \( g(x) = f^{-1}(x) \), we have:
\[ f(g(x)) = x \]
Differentiating both sides with respect to \( x \) gives:
\[ f'(g(x)) \cdot g'(x) = 1 \]
Thus, we can express \( g'(x) \) as:
\[ g'(x) = \frac{1}{f'(g(x))} \]

### Step 2: Find \( g(6) \)
We need to find \( g(6) \), which means we need to find \( f^{-1}(6) \). This requires solving for \( x \) in the equation \( f(x) = 6 \).

### Step 3: Solve \( f(x) = 6 \)
We set up the equation:
\[ 3x^9 - 2x^4 + 2x^3 - 3x^2 + x + \cos x + 5 = 6 \]
This simplifies to:
\[ 3x^9 - 2x^4 + 2x^3 - 3x^2 + x + \cos x - 1 = 0 \]
Testing \( x = 0 \):
\[ f(0) = 3(0)^9 - 2(0)^4 + 2(0)^3 - 3(0)^2 + 0 + \cos(0) + 5 = 0 + 1 + 5 = 6 \]
Thus, \( g(6) = 0 \).

### Step 4: Find \( g'(6) \)
Now we substitute \( g(6) = 0 \) into the expression for \( g'(x) \):
\[ g'(6) = \frac{1}{f'(g(6))} = \frac{1}{f'(0)} \]

### Step 5: Calculate \( f'(x) \)
Now we need to differentiate \( f(x) \):
\[ f'(x) = 27x^8 - 8x^3 + 6x^2 - 6x + 1 - \sin x \]

### Step 6: Evaluate \( f'(0) \)
Substituting \( x = 0 \):
\[ f'(0) = 27(0)^8 - 8(0)^3 + 6(0)^2 - 6(0) + 1 - \sin(0) = 0 + 0 + 0 + 0 + 1 - 0 = 1 \]

### Step 7: Final Calculation of \( g'(6) \)
Now substituting back into our expression for \( g'(6) \):
\[ g'(6) = \frac{1}{f'(0)} = \frac{1}{1} = 1 \]

### Conclusion
Thus, the value of \( g'(6) \) is:
\[ \boxed{1} \]