Define the real-valued function $f: R - \{0\} \rightarrow R$ defined by $f(x) = \frac{1}{x}$,where $x \in R - \{0\}$. Complete the table given below using this definition. What is the domain and range of this function?
$x$ $-2$ $-1.5$ $-1$ $-0.5$ $0.25$ $0.5$ $1$ $1.5$ $2$ $y = \frac{1}{x}$ .... .... .... .... .... .... .... .... ....
| $x$ | $-2$ | $-1.5$ | $-1$ | $-0.5$ | $0.25$ | $0.5$ | $1$ | $1.5$ | $2$ |
| $y = \frac{1}{x}$ | .... | .... | .... | .... | .... | .... | .... | .... | .... |
(N/A) The completed table is given below:
The domain of the function $f(x) = \frac{1}{x}$ is $R - \{0\}$ because the function is undefined at $x = 0$.
The range of the function is $R - \{0\}$ because for any $y \in R - \{0\}$,there exists $x = \frac{1}{y}$ such that $f(x) = y$.
| $x$ | $-2$ | $-1.5$ | $-1$ | $-0.5$ | $0.25$ | $0.5$ | $1$ | $1.5$ | $2$ |
| $y = \frac{1}{x}$ | $-0.5$ | $-0.67$ | $-1$ | $-2$ | $4$ | $2$ | $1$ | $0.67$ | $0.5$ |
The domain of the function $f(x) = \frac{1}{x}$ is $R - \{0\}$ because the function is undefined at $x = 0$.
The range of the function is $R - \{0\}$ because for any $y \in R - \{0\}$,there exists $x = \frac{1}{y}$ such that $f(x) = y$.
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