For the function f(x) = left[ frac{1}{[x]} ri | vedclass

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(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} \right]$.For the function to be defined, the denominator $[x]$ must not be $0$.Thus, $[x]

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The domain is $(-\infty, 0) \cup [1, \infty)$

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(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} ight]$.For the function to be defined, the denominator $[x]$ must not be $0$.Thus, $[x] 
eq 0$, which implies $x Therefore, the domain is $(-\infty, 0) \cup [1, \infty)$.For $x \in [1, 2)$, $[x] = 1$, so $f(x) = [1/1] = 1$.For $x \in [2, \infty)$, $[x] \geq 2$, so $0 For $x \in [-1, 0)$, $[x] = -1$, so $f(x) = [1/(-1)] = -1$.For $x \in [-2, -1)$, $[x] = -2$, so $f(x) = [1/(-2)] = [-0.5] = -1$.Thus, the range is $\{-1, 0, 1\}$.

For the function $f(x) = \left[ \frac{1}{[x]} \right]$, where $[x]$ denotes the greatest integer less than or equal to $x$, which of the following statements is true?

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