Let `g: R to R` be given by g(x)=3 +4x
If `g^(n) (x)=gogo ....og (x)" then "g^(-n)=("where "g^(-n) (x))" denotes inverse of "g^(n) (x))`

Options

A. `(4^(n)-1) +4^(n)x`
B. `(x+1) 4^(-n)-1`
C. `(x+1) 4^(n)-1`
D. `(4^(-n)-1)x+4^(n)`

Answer

To solve the problem, we need to find the inverse of the function \( g^n(x) \), where \( g(x) = 3 + 4x \). We will derive \( g^n(x) \) first and then find its inverse \( g^{-n}(x) \).

### Step-by-Step Solution:

1. **Define the function \( g(x) \)**:
\[ g(x) = 3 + 4x \]

2. **Calculate \( g^2(x) \)**:
\[ g^2(x) = g(g(x)) = g(3 + 4x) \]
Substitute \( 3 + 4x \) into \( g(x) \):
\[ g(3 + 4x) = 3 + 4(3 + 4x) = 3 + 12 + 16x = 15 + 16x \]

3. **Calculate \( g^3(x) \)**:
\[ g^3(x) = g(g^2(x)) = g(15 + 16x) \]
Substitute \( 15 + 16x \) into \( g(x) \):
\[ g(15 + 16x) = 3 + 4(15 + 16x) = 3 + 60 + 64x = 63 + 64x \]

4. **Generalize \( g^n(x) \)**:
From the pattern observed, we can derive a general formula for \( g^n(x) \):
\[ g^n(x) = 3 \cdot (4^{n-1}) + 4^n x \]
This can be expressed as:
\[ g^n(x) = 3 \cdot \frac{4^n - 1}{4 - 1} + 4^n x = \frac{3(4^n - 1)}{3} + 4^n x = 3 + 4^n x \]

5. **Find the inverse \( g^{-n}(x) \)**:
To find the inverse, we set \( y = g^n(x) \):
\[ y = 3 + 4^n x \]
Rearranging for \( x \):
\[ y - 3 = 4^n x \implies x = \frac{y - 3}{4^n} \]
Thus, the inverse function is:
\[ g^{-n}(x) = \frac{x - 3}{4^n} \]

### Final Result:
The inverse of \( g^n(x) \) is:
\[ g^{-n}(x) = \frac{x - 3}{4^n} \]