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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.

Fghj
  • 1,471
  • First draw a picture for the function $\vert x+1 \vert + \vert x+2 \vert + \vert x+3 \vert $, then draw a line which depends on the value $a$. You will see how the number of solutions vary in terms of $a$. – Nirvanacs Aug 04 '17 at 16:41
  • The derivative can be any odd number from $-3$ to $3$, where defined – Akiva Weinberger Aug 04 '17 at 17:28

2 Answers2

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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 $a>2$.
Mark Fischler
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Hint:

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

We have then:

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

JMoravitz
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