Please Explain

If $x\leq3$ , then find the interval between which the value of $\dfrac{1}{x}$ lies.

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $\frac{1}{x} \in (-\infty, 0)\cup [\frac{1}{3}, \infty)$ .

Ameya Daigavane - 5 years, 2 months ago

Thanks a lot...

Sahba Hasan - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$