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My book asks that if $$-5\leq x\leq 1$$ then find the boundaries of absolute value of $x$. Can you please help me in finding that?

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  • $\begingroup$ please define "boundaries of absolute value". Note that some answerers interpreted it differently. $\endgroup$ – Stefan Smith Nov 13 '13 at 1:03
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The midpoint of $[-5,1]$ is $-2$; the distance from $-2$ to both $-5$ and $1$ is $3$. Hence the solution you seek is $$|x-(-2)|\le 3$$ or $$|x+2|\le 3$$

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$$-5\leq x \leq 1$$ $$-5+2\leq x+2 \leq 1+2$$ $$-3\leq x+2 \leq 3$$ $$|x+2|\leq3$$

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  • $\begingroup$ +1. May I ask you to inform me about an ISI maths journal in your erea which covers semigroups or abstract algebra. Thanks. :) $\endgroup$ – mrs Nov 13 '13 at 8:14
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Hint: Find a number $\alpha$ so that \begin{align*} -5 + \alpha &= -\beta \\ 1+\alpha &= +\beta \end{align*} for some number $\beta$. Then $$ -\beta = -5+\alpha \leq x +\alpha \leq 1+\alpha = \beta. $$ That is, you'll obtain $-\beta \leq x+\alpha \leq \beta$. This will give you $|x+\alpha|\leq \beta$.

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I think the replies you received might be a little overwhelming.
1. By definition $|z| \geq 0$ for any $z$ (in other words an absolute value can never be less than 0). Since 0 is within the range of $x$ you were given then 0 is your low boundary.
2. On a (finite) range $[a,b]$ the maximum boundary is the maximum of $-a$ and $b$. In this case $-a$ is$-(-5)$ which is $5$ which is greater then $b$ which is $1$.
So your boundaries are $[0, 5]$

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