1. The problem asks for the range of the function $$f(X) = \frac{2X + 1}{X - 1}$$ and also requests the vertex of this function. 2. First, let's understand the function type: $$f(X)$$ is a rational function, which can be rewritten and analyzed to find its range and vertex. 3. To find the range, set $$y = \frac{2X + 1}{X - 1}$$ and solve for $$X$$ in terms of $$y$$: $$y(X - 1) = 2X + 1$$ $$yX - y = 2X + 1$$ $$yX - 2X = y + 1$$ $$X(y - 2) = y + 1$$ $$X = \frac{y + 1}{y - 2}$$ 4. For $$X$$ to be defined, the denominator $$y - 2 \neq 0$$, so $$y \neq 2$$. 5. Therefore, the range of $$f$$ is all real numbers except $$2$$, which corresponds to option (c) $$\mathbb{R} - \{2\}$$. 6. Now, to find the vertex, rewrite $$f(X)$$ in the form $$a + \frac{b}{X - 1}$$: Divide numerator and denominator: $$f(X) = \frac{2X + 1}{X - 1} = 2 + \frac{3}{X - 1}$$ 7. The function has a vertical asymptote at $$X = 1$$ and a horizontal asymptote at $$y = 2$$. 8. The vertex of this rational function (a hyperbola) is the point where the function changes direction, which occurs at the critical point found by setting the derivative to zero. 9. Compute the derivative: $$f'(X) = \frac{(2)(X - 1) - (2X + 1)(1)}{(X - 1)^2} = \frac{2X - 2 - 2X - 1}{(X - 1)^2} = \frac{-3}{(X - 1)^2}$$ 10. Since $$f'(X) = \frac{-3}{(X - 1)^2} < 0$$ for all $$X \neq 1$$, the function is strictly decreasing on its domain and has no local maxima or minima (no vertex in the usual sense). 11. However, the function's graph is a hyperbola with asymptotes at $$X=1$$ and $$y=2$$, and no vertex point. Final answers: - Range: $$\mathbb{R} - \{2\}$$ (option c) - No vertex exists for this function.