2.1 Radical functions

$\{x\colon x\ge \frac{5}{2}\}()$ since $2x-5\ge 0$ , $2x\ge 5$ , $x\ge \frac{5}{2}$

$\{x\colon x\le 7\}()$ since $7-x\ge 0$ , $-x\ge -7$ , $x\le 7$

$\left\{()\right\}$∞ 2 2 ∞ since $4x^2-16\ge 0$ , $4x^2\ge 16$ , $x^2\ge 4$ , $(x\le -2)\lor (x\ge 2)$

$\left[\frac{-4}{5} , \frac{4}{5}\right]$ since $16-25x^2\ge 0$ , $-25x^2\ge -16$ , $x^2\le \frac{16}{25}$ , $(x\ge \frac{-4}{5})\land (x\le \frac{4}{5})$

$\left\{()\right\}$∞ 1 7 ∞ , $\sqrt{(x-7)(x+1)}\ge 0$

$\left\{()\right\}$∞ 1 2 3 ∞ , $2x^2-7x+3\ge 0$ , $(2x-1)(x-3)\ge 0$ , $(x\le \frac{1}{2})\lor (x\ge 3)$

$\left\{()\right\}$∞ ∞ , since $x^2+4\ge 0$ , $x^2\ge -4$ This will always be true, for all real numbers, any number squared is always positive

$\left\{()\right\}$∞ 8 since $-x+8\ge 0$ , $-x\ge -8$ , $x\le 8$

$\left\{()\right\}$∞ ∞ , since $6x^2+8\ge 0$ , $6x^2\ge -8$ , $x^2\ge \frac{-8}{6}$ This will always be true, for all real numbers, any number squared is always positive

No solution since $-8-6x^2\ge 0$ , $-6x^2\ge 8$ , $x^2\ge \frac{-8}{6}$ This will never be true, so there is no solution, since any number squared is always positive, so it will never be less than 0.