Find all the values of $y$ so that $\min\limits_{[1, 2]}\left | x^{3}- 3x+ y \right |= 6$ .
By Desmos https://www.desmos.com/calculator/i3cesnjguw , I see that the blue line $x= 2$ meets $\left | x^{3}- 3x+ y \right |\leq 6$ at $y_{0}= -8, 4$ , the blue line $x= 1$ meets $\left | x^{3}- 3x+ y \right |\leq 6$ at $y_{0}= -4, 8$ . I guess values of $y$ so that $\min\limits_{[1, 2]}\left | x^{3}- 3x+ y \right |= 6$ are $y= -8, 8$ but why are they ? I tried to use $\min\limits_{[1, 2]}\left | x^{3}- 3x \right |+ \min\limits_{[1, 2]}\left | y \right |\geq\min\limits_{[1, 2]}\left | x^{3}- 3x+ y \right |$ (of course it was an awful idea). I need to the help..
Let $f(x)=x^3-3x$. Note that $f'(x)=3x^2-3$ is positive on the interval $x\in[1,2]$, so $f(x)$ is monotonically increasing. Thus, the minimum value of $|f(x)+y|$ either occurs at $x=1$, or at $x=2$ (because if it were somewhere in the middle, then the minimum value has to be zero, but the question needs it to be $6$). So we are seeking solutions to either $$|f(1)+y|=|y-2|=6$$ or to $$|f(2)+y|=|y+2|=6.$$ Clearly, the solutions are $y=\pm4,\pm8$.
Now we need to check that these values actually work, which I'll leave to you. You should come to the conclusion that, out of these four values, only $y=\pm8$ are solutions.