The problem defines two functions $f(x) = 2x + 3$ and $g(x) = \frac{1}{3}(x - 3)$. We need to find $g^{-1}(x)$ and $f(g^{-1}(x))$.

1. Problem Description

The problem defines two functions $f(x) = 2x + 3$ and $g(x) = \frac{1}{3}(x - 3)$. We need to find $g^{-1}(x)$ and $f(g^{-1}(x))$.

2. Solution Steps

i. Finding $g^{-1}(x)$:

To find the inverse of $g(x)$, we replace $g(x)$ with $y$, swap $x$ and $y$, and solve for $y$.

So we have $y = \frac{1}{3}(x - 3)$.

Swap $x$ and $y$:

$x = \frac{1}{3}(y - 3)$

Multiply both sides by 3:

$3x = y - 3$

Add 3 to both sides:

$y = 3x + 3$

Therefore, $g^{-1}(x) = 3x + 3$.

ii. Finding $f(g^{-1}(x))$:

We need to find $f(g^{-1}(x))$, which means we substitute $g^{-1}(x)$ into $f(x)$.

$f(x) = 2x + 3$

$g^{-1}(x) = 3x + 3$

$f(g^{-1}(x)) = f(3x + 3) = 2(3x + 3) + 3$

$f(g^{-1}(x)) = 6x + 6 + 3$

$f(g^{-1}(x)) = 6x + 9$

3. Final Answer

i. $g^{-1}(x) = 3x + 3$

ii. $f(g^{-1}(x)) = 6x + 9$