A Question About Solving An Equation Involving The Addition Of Absolute Value Functions

Question

I am trying to solve the following problem: $|4-x| \leq |x|-2$. I am trying to do it algebraically, but I'm getting a solution to the problem that makes no sense. I fail to see the error in my reasoning though. I hope to get an explanation where I went wrong.

$|4-x| \leq |x|-2$

$|4-x|-|x| \leq -2$

If $4-x$ and $x$ are both negative, then for them to be equal to $2$, we need to multiply both expressions by $-1$.

$-4+x+x \leq -2$

$-4+2x \leq -2$

$2x \leq 2$

$x \leq 1$

But if you sub in any $x$ less than or equal to $1$, the inequality doesn't work! Can you please explain where in my logic, where in the steps, have I gone wrong? Thank you!

Answer 1

For both $x$ and $4-x$ to be negative, $x<0$ and $x>4$, which is impossible.

As for the solution, the right side has to be positive, hence $| x | \geq 2$. Now, we solve it in $3$ parts.

In $( -\infty,-2 ]$, we have, $4+|x| \leq |x|-2$, which is not true.

In $(4,\infty )$, we have, $x-4 \leq x-2$, which is true.

In $[2,4]$, we have, $4-x \leq x-2 \implies x \geq 3$.

Hence, the solution is, $x \in [3, \infty )$.

Answer 2

hint

observe that there is no solution if the RHS is negative which gives $$|x|<2$$

so, solve in $[2,4], [4,+\infty)$ and $(-\infty,-2]$.

Answer 3

Subtract all terms to the left side of the inequality to obtain

$$ |4-x|-|x|+2\le0 $$

Then define $f(x)=|4-x|-|x|+2$. We want to find all values of $x$ for which $f(x)\le0$.

Important changes occur in the function at $x=0$ and at $x=4$

$$|x|=\begin{cases} x&\text{ for }x\ge0\\-x&\text{ for }x<0\end{cases}$$ $$|4-x|=\begin{cases} 4-x&\text{ for }x\le4\\x-4&\text{ for }x>4\end{cases}$$

This allows us to re-write $f(x)$ as a piecewise defined function: $$f(x)=\begin{cases} 6&\text{ for }x<0\\6-2x&\text{ for }0\le x<4\\-2&\text{ for }x\ge4\end{cases}$$

In the middle interval we see that $f(x)$ is decreasing and reaches a value of $0$ at $x=3$ and we get $f(x)\le0$ for $x\ge3$. So the solution set is $[3,\infty)$.