Let `f:RtoR` be defined as `f(x)=(2x-3pi)^(3)+(4)/(3)x+cosx` and `g(x)=f^(-1)(x)` then

Options

A. `g^(')(2pi)=(3)/(7)`
B. `g^(')(2pi)=(7)/(3)`
C. `g^('')(2pi)=0`
D. `g^('')(2pi)=-(27)/(243)`

Explanation

To solve the problem step by step, we need to find the value of \( g(2\pi) \) where \( g(x) = f^{-1}(x) \) and \( f(x) = (2x - 3\pi)^3 + \frac{4}{3}x + \cos x \).

### Step 1: Find \( x \) such that \( f(x) = 2\pi \)

We need to find \( x \) such that \( f(x) = 2\pi \). This means we need to solve the equation:

\[ (2x - 3\pi)^3 + \frac{4}{3}x + \cos x = 2\pi \]