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How to use the absolute value function to translate each of the following statements into a single inequality.

(a) $\ x ∈ (-4,10) $

(b) $\ x ∈ (-\infty,2] \cup[9,\infty) $

I think in the first one the absolute value of $\ x$ should be greater than 4 and less than 10. is that correct? because the distance from $\ x$ to $\ 0$ should be between $\ 4$ and$\ 10$ in order for $\ x$ to belong in this interval.

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a) Since $3$ is in the midlle of $(-4,10)$ we see that $x$ is at most $7$ (but not equal) distance from $3$, so $$|x-3|<7$$

b) Since $x$ is at least ${7\over 2}$ from ${11\over 2}$ we have $$|x-{11\over 2}|\geq {7\over 2}$$

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  • $\begingroup$ can u please explain what's is case if the interval has a value instead of negative infinity and infinity in b $\endgroup$ – John C. Dec 15 '19 at 22:34
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$x \in (-4,10)\implies$

$-4 < x < 10\implies$

$-4 + k < x + k < 10 + k$

If we have $M=|10+k| = |-4+k|$ (and presumably $-M= -4k < 0 < 10+k=M$) then we would have $-M < x + k < M$ so $|x + k| <M$.

So if $10+k = -(-4+k)=4-k$ then $2k = -6$ and $k -3$ and $M=10 -3=7$ and

$-4 -3 < x - 3< 10-3$

$-7 < x-3 < 7$

$|x-3| < 7$.

b) $x ∈ (-\infty,2] \cup[9,\infty)$ means

$x \le 2$ or $x \ge 9$.

Again $x + k \le 2+k$ or $x \ge 9+k$. If we can get $M=9+k = -(2+k)$ we would have $|x+k | \ge M$.

$9+k = -(2+k) \implies k=-5.5$ so $x-5.5 \le -3.5$ and $x-5.5 \ge 3.5$ so $|x-5.5| \ge 3.5$.

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