If f(x) = 3 - x^2 for 1 le x le 4,then the do | vedclass

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(D) Given $f(x) = 3 - x^2$ where $1 \le x \le 4$.For the function $f(2x)$,we substitute $2x$ in place of $x$:$f(2x) = 3 - (2x)^2 = 3 - 4x^2$.The condi

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$[\frac{1}{2}, \frac{\sqrt{3}}{2})$

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(D) Given $f(x) = 3 - x^2$ where $1 \le x \le 4$.For the function $f(2x)$,we substitute $2x$ in place of $x$:$f(2x) = 3 - (2x)^2 = 3 - 4x^2$.The condition $1 \le x \le 4$ for $f(x)$ implies that for $f(2x)$,the argument $2x$ must satisfy $1 \le 2x \le 4$,which simplifies to $\frac{1}{2} \le x \le 2$.For the logarithmic function $\log_e(f(2x))$ to be defined,the argument must be strictly positive:$f(2x) > 0 \implies 3 - 4x^2 > 0$.$4x^2 Since $x$ must be positive from the domain constraint,we have $0 \le x Combining the conditions $\frac{1}{2} \le x \le 2$ and $x $\frac{1}{2} \le x Thus,the domain is $[\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

If $f(x) = 3 - x^2$ for $1 \le x \le 4$,then the domain of $\log_e(f(2x))$ is:

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