Question

The domain of definition of the function $f(x)=\sqrt{2^{2 x}+(64)^{\frac{(x-2)}{3}}-2^{-1}\left(72+2^{2 x}\right)}$ is

• A. $-\infty, \infty$
• B. $-\infty,-3$
• C. $3, \infty$
• D. $19,1$

Answer 1

Answer: (C)

For $f(x)$ to be defined,
$2^{2 x}+(64)^{\frac{(x-2)}{3}}-2^{-1}\left(72+2^{2 x}\right) \geq 0 $
$\Rightarrow 4^{x}+4^{x-2}-36-\frac{4^{x}}{2} \geq 0 $
$\Rightarrow y+\frac{y}{16}-36-\frac{y}{2} \geq 0, \text { where } y=4^{x} $
$\Rightarrow y \geq 64 \Rightarrow 4^{x} \geq 4^{3} \Rightarrow x \geq 3$
$\therefore \text { Domain of } f=[3, \infty]$