What are the number solutions of this equation

Question

I am new to modulus function, and I want to find number of solutions to:

$$ \vert x+1 \vert + \vert x+2 \vert + \vert x+3 \vert = a $$

$$ x = [-4,4] $$

Where $ a $ is a parameter.

Answer 1

Let $y=x+2$. Then define $f(y)$ by $$ |x+1|+|x+2|+|x+3| = |y-1|+|y|+|y+1| \equiv f(y) $$ $f(y) = |1-y|+|y|+|1+y|$ so $f(-y) = f(y)$.

Above $y=1$, $f(y) = 3y$ which is monotonic increasing. For $0\leq y \leq 1$, $$ f(y) = (1-y)+y+(1+y) = 2+y $$ which is also monotonic increasing and has a minimum at $f(0)=2$.

So $f(y)$ has a minimum value of $2$ at $y=0$, and is increasing for positive $y$ and decreasing for negative $y$. Therefore, $f(y)=a$ (and therefore your original equation) has:

• No solutions for $a<2$.
• One solution for $a=2$.
• Two solutions for $a>2$.

Answer 2

Hint:

$|x|=\begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$

We have then:

$$|x+1|+|x+2|+|x+3|=\begin{cases}x+1+x+2+x+3&\text{if}~x\geq -1\\ -(x+1)+x+2+x+3&\text{if}~-2\leq x<-1\\\vdots&\text{if}-3\leq x<-2\\\vdots&\text{if}~x<-3\end{cases}$$