Let f(x) = (2 x + 1)/(x - 3) for every real x ≠ 3.

(a) Inverse. From $y = \dfrac{2x + 1}{x - 3}$: $xy - 3y = 2x + 1$, so $x = \dfrac{3y + 1}{y - 2}$ and $f^{-1}(x) = \frac{3 x + 1}{x - 2}$. Then $f^{-1}(5) = \frac{16}{3} = 5.33$.

(b) Composition. $f(4) = 9$, then $f(9) = \frac{19}{6} = 3.17$.

(c) Range. $x = \dfrac{3y + 1}{y - 2}$ has a solution for every $y$ except $y = 2$ (then $x(y - 2) = 3y + 1$ reads $0 = 7$, which is impossible). So $2$ is the one value $f$ never takes.

(d) Domain of $f \circ f$. We need $x \ne 3$ and $f(x) \ne 3$. $\dfrac{2x + 1}{x - 3} = 3$ gives $2x + 1 = 3x - 9$, so $x = 10$. Indeed $f(f(x)) = - \frac{5 x - 1}{x - 10}$ breaks down at $x = 10$.

(e) Fixed points. $2x + 1 = x(x - 3)$ gives $x^{2} - 5 x - 1 = 0$, with roots $\frac{5}{2} - \frac{\sqrt{29}}{2} \text{ and } \frac{5}{2} + \frac{\sqrt{29}}{2}$. The larger is $\frac{5}{2} + \frac{\sqrt{29}}{2} = 5.19$.