1. We are given the inequality involving a function $f(x)$: $$\sqrt{x} + 7 \leq f(x) \leq \frac{x - 1}{2}.$$\n\n2. This inequality states that the value of the function $f(x)$ is bounded below by $\sqrt x + 7$ and bounded above by $\frac{x-1}{2}$.\n\n3. To understand the domain where this inequality can hold, we consider the domain restrictions and the ordering of bounds:\n\n- Since $\sqrt{x}$ is defined only for $x \geq 0,$ the domain of $x$ must satisfy $x \geq 0$.\n\n- For $f(x)$ to exist and satisfy both inequalities, the lower bound must be less than or equal to the upper bound:\n$$\sqrt{x} + 7 \leq \frac{x - 1}{2}.$$\n\n4. Solve the inequality for $x$:\nMultiply both sides by 2 to clear the denominator:\n$$2\sqrt{x} + 14 \leq x - 1.$$\nRearranged:\n$$x - 1 - 2\sqrt{x} - 14 \geq 0 \Rightarrow x - 2\sqrt{x} - 15 \geq 0.$$\n\n5. Substitute $t = \sqrt{x}$ with $t \geq 0$, so $x = t^2$:\n$$t^2 - 2t - 15 \geq 0.$$\n\n6. Factor or solve the quadratic inequality:\nThe quadratic equation $t^2 - 2t - 15 = 0$ has roots from \n$$t = \frac{2 \pm \sqrt{4 + 60}}{2} = 1 \pm 4 = 5 \text{ or } -3.$$\nSince $t \geq 0$, we discard $t = -3$.\n\n7. The inequality $t^2 - 2t - 15 \geq 0$ holds for $t \leq -3$ or $t \geq 5$, but only $t \geq 5$ is valid for $t \geq 0$.\n\n8. Substitute back $t = \sqrt{x}$:\n$$\sqrt{x} \geq 5 \Rightarrow x \geq 25.$$\n\n9. Combine this with the domain $x \geq 0$, the solution interval where the inequality makes sense is $$x \geq 25.$$\n\n10. Hence, for $x \geq 25$, the function $f(x)$ lies between the lower bound $\sqrt{x} + 7$ and the upper bound $\frac{x-1}{2}$.