A unique solution

Question

Find the sum of all values of k so that the system $$y=|x+23|+|x−5|+|x−48|$$$$y=2x+k$$ has exactly one solution in real numbers. If the system has one solution, then one of the three $x's$, should be 0. Then, there are 3 solutions, when each of the modulus becomes 0. Where am I going wrong?

Answer 1

Perhaps it would help to write the first equation as $$ y= \begin{cases} -3x+30&\text{ for }x\leq-23\\ -x+76&\text{ for }x\in[-23,5]\\ x+66&\text{ for }x\in[5,48]\\ 3x-30&\text{ for }x\geq48\\ \end{cases} $$ Then find $x$-values of intersections with each of those four lines: $$ \begin{align} 2x+k&=-3x+30&&\iff x=\frac{30-k}5\\ 2x+k&=-x+76&&\iff x=\frac{76-k}{3}\\ 2x+k&=x+66&&\iff x=66-k\\ 2x+k&=3x-30&&\iff x=30+k \end{align} $$ and then write $k$-ranges for the valid intersections: $$ \begin{align} \frac{30-k}{5}&\leq-23&&\iff k\geq 30+5\cdot 23 =145\\ \frac{76-k}{3}&\in[-23,5]&&\iff k\in[76-3\cdot 5,76+3\cdot 23]=[61,145]\\ 66-k&\in[5,48]&&\iff k\in[18,61]\\ 30+k&\geq 48&&\iff k\geq 18 \end{align} $$ So whenever $k$ is in one of the following four intervals, $y=2x+k$ has a legal intersection with the corresponding segment of $y=|x+23|+|x-5|+|x-48|$: $$ \begin{align} -3x+30:&k\in[145,\infty)\\ -x+76:&k\in[61,145]\\ x+66:&k\in[18,61]\\ 3x-30:&k\in[18,\infty) \end{align} $$ and we see that only for $k=18$ we have $y=2x+k$ intersecting $y=x+66$ and $y=3x-30$ at the same point, $x=66-18=30+18=48$.

---

Here is a diagram of the situtation:

Answer 2

Working with String's initial boundaries, we have $$ 0= \begin{cases} -5x+30-k&\text{ for }x\leq-23\\ -3x+76-k&\text{ for }x\in[-23,5]\\ -x+66-k&\text{ for }x\in[5,48]\\ x-30-k&\text{ for }x\geq48\\ \end{cases} $$ Now notice that the function is decreasing for $x<48$ and increasing for $x>48$. This means that the function $f(x)=y-2x-k$ has a global minimum at $x=48$. Since the function is continuous (it is a sum of absolute value functions and a linear function which are each continuous), when $k=18$, $x=48$ is the unique solution of $f(x)=0$; when $k>18$, the function has two roots; when $k<18$, the function has none. Hence the only time when there is a unique solution is when $k=18$. To picture why this is, think of the curve $y=x^2-k$ and its roots.

Answer 3

$$y-2x$$ is a piecewise linear function and is convex. The slope is nowhere zero (slope values are $-5,-3,-1,1$) so that it achieves an isolated minimum, which is the requested value of $k$, at the lowest vertex.

$$\begin{align}-23&\to145\\5&\to61\\48&\to\color{green}{18}\end{align}$$