Respuesta :

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]

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