The range of the function f(x) = frac{x + 2}{ | vedclass

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(B) Given the function $f(x) = \frac{x + 2}{|x + 2|}$.We know that the absolute value function $|x + 2|$ is defined as:$|x + 2| = \begin{cases} x + 2,

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$\{-1, 1\}$

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(B) Given the function $f(x) = \frac{x + 2}{|x + 2|}$.We know that the absolute value function $|x + 2|$ is defined as:$|x + 2| = \begin{cases} x + 2, & 	ext{if } x + 2 > 0 	ext{ (i.e., } x > -2) \ -(x + 2), & 	ext{if } x + 2 For $x > -2$,$f(x) = \frac{x + 2}{x + 2} = 1$.For $x Note that the function is undefined at $x = -2$ because the denominator becomes zero.Therefore,the range of the function $f(x)$ is $\{-1, 1\}$.

The range of the function $f(x) = \frac{x + 2}{|x + 2|}$ is

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