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If $|x + 3| < 0.5$, show that $|4x + 13| < 3$

This is what I've got so far:

$|4x + 13| = |(x + 3) + (3x + 10)|$

by the Triangle Inequality:

$|(x + 3) + (3x + 10)| \le |x + 3| + |3x + 10|$

Now I continue to apply the Triangle Inequality to reach:

$|(x + 3) + (3x + 10)| \le |x + 3| + |x + 3| + |x + 3| + |x + 3| + |1|$

So I come to the conclusion: $|4x + 13| \le 4|x + 3| + 1$

Since $|x + 3| < 0.5$, then $4|x + 3| + 1 < 4\cdot 0.5 + 1 = 3$

Then $|4x + 13| < 3$

Please take a look and let me know if there are errors, if so, enlighten me.

Thank you :)

jojeck
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Kermit the Hermit
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3 Answers3

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You can also remove the absolute value and work with a two-sided equation. It's a little simpler, as it doesn't involve the triangle inequality.

$$-0.5<x+3<0.5$$ $$-2<4(x+3)<2$$ $$-2<4x+12<2$$ $$-1<4x+13<3$$ Therefore: $$|4x+13|<3$$

Duncan
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Yes, your solution is correct. A quicker approach could go like this: $|4x+13|=|4(x+3)+1|\leq 4|x+3|+1<4\cdot\frac12+1=3$.

Alraxite
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The function $ g(x) = | 4 x +13|$ is convex. The inequality $|x+3| < 1/2$ says we should choose as a domain $ I = [ -7/2 ,-5/2]$. Now since $g$ is convex it takes it's maximum at the boundary so it is either $ g(-5/2)= 3$ or $ g(-7/2)= 1$. Hence the maximum of $g$ is $3$ on $I$.

But since $g$ is non constant on $I$ it does not attain it's maximum anywhere inside the $I$, which translates to $ g(x) < 3$ for all $x$ such that $ |x+3|<1/2$.

clark
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