We are given a function $g(x) = \frac{2x+3}{3x-1}$ defined for $x \neq \frac{1}{3}$. We need to find (i) the image of 2 under the function $g$, i.e., $g(2)$, and (ii) the inverse function $g^{-1}(x)$ and the value of $g^{-1}(\frac{1}{2})$.

1. Problem Description

We are given a function $g(x) = \frac{2x+3}{3x-1}$ defined for $x \neq \frac{1}{3}$.

We need to find (i) the image of 2 under the function $g$, i.e., $g(2)$, and (ii) the inverse function $g^{-1}(x)$ and the value of $g^{-1}(\frac{1}{2})$.

2. Solution Steps

(i) To find $g(2)$, substitute $x=2$ into the expression for $g(x)$:

$g(2) = \frac{2(2)+3}{3(2)-1} = \frac{4+3}{6-1} = \frac{7}{5}$

(ii) To find the inverse function $g^{-1}(x)$, let $y = g(x)$. Then $y = \frac{2x+3}{3x-1}$.

To find the inverse, we swap $x$ and $y$ and solve for $y$:

$x = \frac{2y+3}{3y-1}$

$x(3y-1) = 2y+3$

$3xy - x = 2y + 3$

$3xy - 2y = x+3$

$y(3x-2) = x+3$

$y = \frac{x+3}{3x-2}$

Therefore, $g^{-1}(x) = \frac{x+3}{3x-2}$.

Now we need to find $g^{-1}(\frac{1}{2})$. Substitute $x = \frac{1}{2}$ into the expression for $g^{-1}(x)$:

$g^{-1}(\frac{1}{2}) = \frac{\frac{1}{2} + 3}{3(\frac{1}{2})-2} = \frac{\frac{1}{2} + \frac{6}{2}}{\frac{3}{2} - \frac{4}{2}} = \frac{\frac{7}{2}}{-\frac{1}{2}} = \frac{7}{2} \cdot \frac{2}{-1} = -7$

3. Final Answer

(i) $g(2) = \frac{7}{5}$

(ii) $g^{-1}(x) = \frac{x+3}{3x-2}$ and $g^{-1}(\frac{1}{2}) = -7$