The range of the function f : R rightarrow R, | vedclass

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(C) To find the range of $f(x) = \frac{(x + 1)^4}{x^4 + 1}$,we observe the following:$1$. Since $(x+1)^4 \geq 0$ and $x^4 + 1 > 0$ for all $x \in R$,t

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$[0, 8]$

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(C) To find the range of $f(x) = \frac{(x + 1)^4}{x^4 + 1}$,we observe the following:$1$. Since $(x+1)^4 \geq 0$ and $x^4 + 1 > 0$ for all $x \in R$,the minimum value of $f(x)$ is $0$ (at $x = -1$).$2$. To find the maximum value,we use the power mean inequality or Jensen's inequality. For the convex function $g(t) = t^4$,by Jensen's inequality:$\frac{g(x) + g(1)}{2} \geq g\left(\frac{x+1}{2}\right)$$\frac{x^4 + 1}{2} \geq \left(\frac{x+1}{2}\right)^4 = \frac{(x+1)^4}{16}$$3$. Rearranging the inequality:$\frac{(x+1)^4}{x^4 + 1} \leq \frac{16}{2} = 8$$4$. Thus,the range of the function is $[0, 8]$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

The range of the function $f : R \rightarrow R$,defined by $f(x) = \frac{(x + 1)^4}{x^4 + 1}$,is

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$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
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