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The question below is in my textbook about ınequalities of absolute values. The question Express the interval in terms of an inequality involving absolute value. is asked. I do not understand how the answers correlate with the questions above. How would I be able to get the answers for them?

Examples:

1.[-2, 2]
2.(0,4)
3.[-1, 8]

Answers:

1. |x| ≤ 2  
2. |x - 2| < 2
3. |x - 3.5| ≤ 4.5
Master Selcuk
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2 Answers2

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  1. $$|x+3|\leq7\tag{*}$$ literally reads as $$\text{“the absolute value of $(x+3)$ is less than or equal to $7$”},$$ which means $$\text{“when the sign of $(x+3)$ is disregarded, its value is at most $7$”}.$$

    There are exactly two possibilities for $(x+3)$:

    • if it is nonnegative, then clearly $0\leq (x+3)\leq7$;
    • if it is negative, then, noting that $(-\infty,-7)$ is inadmissible, $-7\leq(x+3)<0.$

    In other words, $-7\leq(x+3)\leq7.\quad$ (If $x$ belongs to either set $A$ or $B,$ then it belongs to set $A\cup B.)$

    In general, each of your three given exercises can be handled by first applying this definitionҗ literally to the LHS of the inequality: \begin{align}|x| &= \begin{cases}-x &\text{ if }x<0; \\x &\text{ if }x\geq0\end{cases}\end{align} җCultivating the habit of reflexively turning to definitions is extremely valuable for doing Mathematics!

  2. Alternatively, by separately plotting $y=|x+3|$ and $y=7,$

    enter image description here

    we can see that the interval for which the ineqaulity $(*)$ is true is $[-10,-4].$

    (The graph of $y=f(x)$ is simply the graph of $y=f(x)$ but with any negative-$y$ portion reflected in the $x$-axis.)

ryang
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Given any interval $(a,b)$, we have $(a,b)$ = $\left(\frac{a+b}2-\frac{b-a}2,\frac{a+b}2+\frac{b-a}2\right)$ (this can be verified with simply algebra). If we define $c = \frac{a+b}2$, $r = \frac{b-a}2$, then the interval is $(c-r,c+r)$. If we consider numbers of the form $c+d$, then the interval is the set of all values for which $|d|<r$. You can think of the interval as a one-dimensional version of a circle; it's the set of all points within a distance $r$ of the center $c$. We find the center by taking the midpoint (or average) of the two bounds, and we find the radius by taking half the difference (you can think of the full difference as the "diameter" of the "circle", so the radius is half of that).

Acccumulation
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