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If $x$ is a real number, then $|x+1| \leq 3$ implies that $-4 \leq x \leq 2$.

I've tried to prove this by exhaustion, is that the right way to prove it?

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    $\begingroup$ Yes, you have $-3 \leq x + 1 \leq 3$ $\endgroup$ – GAVD Aug 31 '17 at 6:03
  • $\begingroup$ (Not recommended in this case, but...) For a different approach: $\;|x+1|\le 3 \iff (x+1)^2-9 \le 0$ $\iff (x+4)(x-2) \le 0$ $\iff -4 \le x \le 2\,$. $\endgroup$ – dxiv Aug 31 '17 at 6:10
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If $a$ and $b$ are real numbers and if $b \ge 0$, then the inequality $|a| \le b$ exactly means

$$-b \le a \le b.$$

Hence , if $b=3$, then

$$|a| \le 3 \iff -3 \le a \le 3.$$

Now let $a=x+1$. Then we get

$$|x+1| \le 3 \iff -3 \le x+1 \le 3 \iff -4 \le x \le 2.$$

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You can use the definition of $|.|$.

The absolute value of real number $a$ (we'll write it $|a|$) it's the distance between the point,

which corresponds to $a$ in the $x$-axis and the origin.

In our case the distance between $x+1$ and $0$ less or equal to $3$.

Thus, we have $$-3\leq x+1\leq3$$ or $$-4\leq x\leq2$$

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For real valued $x$ we have two cases.

  • Either $x+1\geq 0$, in which case $\lvert x+1\rvert =(x+1)$ so... $0\leq x+1 \leq 3$

  • Or $x+1<0$, in which case $\lvert x+1\rvert=-(x+1)$, so... $0 < -(x+1)\leq 3$.

Thus $(\lvert x+1\rvert \leq 3 )\iff ( -3\leq x+1 \leq 3 )$

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A good thing to think about expressions like $| x+ a| \leq b $ is that they are equivalent to saying $|x - (-a)| \leq b$, or that the 'distance' between $x$ and $-a$ is less than $b$. Then in the context of your problem it is clear that $x$ lies between the bounds given.

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I'd say that the easiest way to prove this is to simply solve the inequality for x.

We know that:

|x+1| = 3, simplifies to:

x+1 = +3 AND x+1=-3

So by this logic, we can say that:

x+1 ≤ 3 AND x+1 ≥ -3 (switching the ≤ to ≥ when we change the sign on the 3)

x ≤ 2 AND x ≥ -4 (subtracting 1 in on both sides)

Then we can just simplify this expression down into the one, written as, -4 ≤ x ≤ 2, just as you needed to show.

Hope this helps. This is just the easiest way to show it, though there are many others.

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  • $\begingroup$ The statement $|x+ 1| \leq 3$ is equivalent to the statement $x + 1 \leq 3$ AND $x + 1 \geq -3$. Every real number satisfies $x + 1 \leq 3$ OR $x + 1 \geq -3$. $\endgroup$ – N. F. Taussig Aug 31 '17 at 11:27

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