Suppose \( x > 0 \).
Then \( x \leq 3 \Rightarrow \frac{1}{x} \geq \frac{1}{3} \) (on dividing by x).
This means \(\frac{1}{x} \in [\frac{1}{3}, \infty)\).

If \( x < 0 \) then \(\frac{1}{x} \in (-\infty, 0)\). (Look at the graph!).
\(\frac{1}{x}\) is not defined at \( x = 0\).

Combining all this we have \(\frac{1}{x} \in (-\infty, 0)\cup [\frac{1}{3}, \infty)\).

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TopNewestSuppose \( x > 0 \).

Then \( x \leq 3 \Rightarrow \frac{1}{x} \geq \frac{1}{3} \) (on dividing by x).

This means \(\frac{1}{x} \in [\frac{1}{3}, \infty)\).

If \( x < 0 \) then \(\frac{1}{x} \in (-\infty, 0)\). (Look at the graph!).

\(\frac{1}{x}\) is not defined at \( x = 0\).

Combining all this we have \(\frac{1}{x} \in (-\infty, 0)\cup [\frac{1}{3}, \infty)\).

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Thanks a lot...

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