If $$f(x) = x^2 - x + 5$$, $$x > \frac{1}{2}$$, and g(x) is its inverse function, then g'(7) equals:

We are given the function $$ f(x) = x^2 - x + 5 $$ for $$ x > \frac{1}{2} $$, and we know that $$ g(x) $$ is its inverse function. We need to find $$ g'(7) $$.

To find the derivative of the inverse function at a point, we use the formula:

$$ g'(y) = \frac{1}{f'(g(y))} $$

So, for $$ y = 7 $$, we have:

$$ g'(7) = \frac{1}{f'(g(7))} $$

This means we need to find two things: first, the value of $$ g(7) $$, which is the $$ x $$ such that $$ f(x) = 7 $$, and second, the derivative $$ f'(x) $$ evaluated at that $$ x $$.

Start by solving $$ f(x) = 7 $$:

$$ x^2 - x + 5 = 7 $$

Subtract 7 from both sides:

$$ x^2 - x + 5 - 7 = 0 $$

$$ x^2 - x - 2 = 0 $$

Now, solve this quadratic equation. We factor it:

$$ x^2 - x - 2 = (x - 2)(x + 1) = 0 $$

So, the solutions are $$ x = 2 $$ and $$ x = -1 $$.

But the domain is restricted to $$ x > \frac{1}{2} $$. Since $$ -1 < \frac{1}{2} $$, we discard $$ x = -1 $$. Thus, $$ x = 2 $$ is the solution.

Therefore, $$ g(7) = 2 $$.

Next, find the derivative of $$ f(x) $$. The function is $$ f(x) = x^2 - x + 5 $$, so:

$$ f'(x) = \frac{d}{dx}(x^2 - x + 5) = 2x - 1 $$

Now, evaluate $$ f'(x) $$ at $$ x = g(7) = 2 $$:

$$ f'(2) = 2(2) - 1 = 4 - 1 = 3 $$

Finally, substitute back into the inverse derivative formula:

$$ g'(7) = \frac{1}{f'(g(7))} = \frac{1}{f'(2)} = \frac{1}{3} $$

So, $$ g'(7) = \frac{1}{3} $$.

Looking at the options:

A. $$ -\frac{1}{3} $$

B. $$ \frac{1}{13} $$

C. $$ \frac{1}{3} $$

D. $$ -\frac{1}{13} $$

Hence, the correct answer is Option C.