Solving $2|x+1|>|x+4|$

Question

I'm trying to solve the following equations and inequalities for $x\in\mathbb R$:

$$2|x+1|>|x+4|$$

I know I'm supposed to consider the intervals $(-\infty,-4), [-4,-1]$ and $(-1,\infty)$ but don't know how each affects the equation and how to conclude a result.

I think considering interval $(-\infty,-4)$: $2|1+x|>|4+x|$ $\Rightarrow x>2$ but if I'm right how will it work for the other intervals?

Answer 1

$$\begin{align} &4(x+1)^2>(x+4)^2\\ &4x^2+8x+4>x^2+8x+16\\ &3x^2-12>0\\ &x>2\ \text{ or }\ x<-2 \end{align}$$

Answer 2

The considered intervals are right, use the definition of the absolute value $$|x|=\left\{ \begin{array}{rl} x & x\geq 0 \\ -x & x < 0 \\ \end{array}\right. $$ In the Intervalls you can always think which one is the case and so eliminate the absolut value.

To check your results here is a plot, the red one is $2\cdot |x+1|$

For $x \in (-4,-1)$ we have $x+4 >0$ and $x+1<0$

So the inequality is \begin{align*} 2 |x+1| & > |x+4| \\ -2\cdot (x+1) &> x+4 \\ \iff -2-2x &> x+4 \\ \iff -6 &> 3 x \\ \iff -2 &> x \end{align*}

Answer 3

What is the sign of $x+1$ and $x+4$ in each interval?

Then use $|y|=y$ if $y\geq 0$, and $|y|=-y$ if $y\leq 0$.

For instance, if $x\in [-4,-1]$, you have $x+4\geq 0$ so $|x+4|=x+4$ and $x+1\leq 0$ so $|x+1|=-x-1$. In this interval: $$ 2|x+1|>|x+4|\quad\Leftrightarrow\quad 2(-x-1)>x+4\quad\Leftrightarrow\quad 3x<-6\quad\Leftrightarrow\quad x<-2. $$ So this yields $[-4,-2)$.

Repeat the procedure on the other intervals.

Alternative: draw simultaneously the graphs of $2|x+1|$ and $|x+4|$. Then find where the former is above the latter. It is very easy. You just have to find the two intersections.

See here.