$x$ and $y$ are real numbers such that $y = |x - 2| - |2x - 12| + |x - 8|$. What is the least possible value of $y$?

Question

I solved this question by assuming cases for various roots of this equation and I found that lowest value of $y$ is $-2$ for $x<2$ but this is a long process and took some time. I was wondering if there is a quick way to solve this or any smart trick to get the answer while solving these types of questions in the competitive exams. Please help !!!

Thanks in advance !!!

Answer 1

Note that the function will be linear in each of the subintervals determined by the $x$-values $2$, $6$, $8$. Evaluate the function at each of these three $x$-values; and at one point to the left of $2$ and at one point to the right of $8$, and you should be able to resolve the question.

Answer 2

That is a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is differentiable everywhere except the points $\{2,6,8\}$.
Also it is obvious that for $x<2$ and $x>8$, $f$ is constant.
It should be easy to show that its minimum value is $-2$ and its maximum is $6$. You could simply "break" the line of reals into disjoint subintervals in order to "break" the form of $f$ into more simple forms without absolute values. Give it a try.