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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 !!!

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  • $\begingroup$ @MyMolecules : Commented on that answer. Sorry I overlooked that notifications for that. $\endgroup$
    – Ganit
    Commented Dec 11, 2021 at 15:55

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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.

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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.

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