Answer:
[tex]0\le x\le 64[/tex]
Step-by-step explanation:
[tex]\dfrac{1}{2\sqrt{x} } \geq \dfrac{1}{16}[/tex]
Using substitution
[tex]\sqrt{x} = t[/tex]
[tex]\dfrac{1}{2t} \geq \dfrac{1}{16}[/tex]
Multiply both sides by 2
[tex]\dfrac{1}{t} \geq \dfrac{1}{8}[/tex]
Multiply both sides by [tex]t[/tex]. We can do it because [tex]t > 0[/tex]
[tex]1 \geq \dfrac{t}{8}[/tex]
Multiply both sides by 8
[tex]8 \geq t[/tex]
As [tex]t=\sqrt{x}[/tex]
[tex]8 \geq \sqrt{x}[/tex]
So
[tex]x\le 64\cap x\ge 0[/tex]
[tex]0\le x\le 64[/tex]