Question

$\frac{1}{2 \sqrt{x} } \geqslant \frac{1}{16}$
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Answer 1

Answer:

$0\le x\le 64$

Step-by-step explanation:

$\dfrac{1}{2\sqrt{x} } \geq \dfrac{1}{16}$

Using substitution

$\sqrt{x} = t$

$\dfrac{1}{2t} \geq \dfrac{1}{16}$

Multiply both sides by 2

$\dfrac{1}{t} \geq \dfrac{1}{8}$

Multiply both sides by $t$. We can do it because $t > 0$

$1 \geq \dfrac{t}{8}$

Multiply both sides by 8

$8 \geq t$

As  $t=\sqrt{x}$

$8 \geq \sqrt{x}$

So

$x\le 64\cap x\ge 0$

$0\le x\le 64$