Prove $|x+1|\leq 4$ implies that $-4\leq x\leq 2$.

Question

How do I prove that if $x$ is a real number, then $\lvert x+1 \rvert\leq 3$ implies that $-4\leq x\leq 2$.

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EDIT: $\lvert x+1 \rvert\leq 4$ should be $\lvert x+1 \rvert\leq 3$

Answer 1

When one says $|x+a|\leq b$ for any $a\in \mathbb{R}$ and $b\ge 0$, it means that $x+a\leq b$ or $-(x+a)\leq b \Leftrightarrow x+a\geq -b$, which in turns is the same as saying $$ -b\leq x+a\leq b\\ -b-a\leq x\leq b-a. $$ In your case, $a=1$ and $b=3$, and so $$ -3-1\leq x\leq 3-1,\\ -4\leq x\leq 2 $$

Answer 2

One definition for $\;|\cdot|\;$ is $$|x| = x \max -x$$ Using this definition, and a basic property of $\;\max\;$, we can simply calculate \begin{align} & |x+1| \leq 3 \\ \equiv & \;\;\;\;\;\text{"the above definition for $\;|\cdot|\;$"} \\ & (x+1) \max -(x+1) \leq 3 \\ \equiv & \;\;\;\;\;\text{"basic property of $\;\max\;$"} \\ & x+1 \leq 3 \;\land\; -(x+1) \leq 3 \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & -4 \leq x \leq 2 \\ \end{align} So we see that actually both statements are equivalent.