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.
On
$x \in (-4,10)\implies$
$-4 < x < 10\implies$
$-4 + k < x + k < 10 + k$
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 $|x-5.5| \ge 3.5$.
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 $$|x-{11\over 2}|\geq {7\over 2}$$