Question Number 157604 by mnjuly1970 last updated on 25/Oct/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Q}{uestion}\: \\ $$$$\:{find}\:{the}''\:{minimum}''\:{value}\:{of}: \\ $$$$ \\ $$$${f}\:\left({x}\right):=\:\mid\mathrm{1}+{x}\mid+\mid\:\mathrm{2}+{x}\mid\:+\:\mid\mathrm{4}\:+\mathrm{2}{x}\mid \\ $$$$ \\ $$
Commented by mr W last updated on 25/Oct/21
$${i}\:{think}\:{the}\:{easiest}\:{way}\:{is}\:{to}\:{check}\:{all} \\ $$$${kink}\:{positions}: \\ $$$${x}=−\mathrm{1},\:−\mathrm{2} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{0}+\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{1}+\mathrm{0}+\mathrm{0}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)_{{min}} =\mathrm{1} \\ $$
Commented by mr W last updated on 25/Oct/21
$${in}\:{this}\:{case}\:{we}\:{can}\:{also}\:{do}\:{like}\:{this}: \\ $$$${f}\left({x}\right)=\mid\mathrm{1}+{x}\mid+\mathrm{3}\mid\mathrm{1}+\mathrm{1}+{x}\mid=\mid\mathrm{1}+{x}\mid+\mathrm{3}\pm\mathrm{3}\mid\mathrm{1}+{x}\mid \\ $$$$=\mathrm{3}+\begin{cases}{\mathrm{4}\mid\mathrm{1}+{x}\mid}\\{−\mathrm{2}\mid\mathrm{1}+{x}\mid}\end{cases} \\ $$$${f}\left({x}\right)_{{min}} =\mathrm{3}−\mathrm{2}=\mathrm{1}\:{at}\:{x}=−\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 25/Oct/21
$${grateful}\:{sir}\:{W} \\ $$
Answered by TheSupreme last updated on 25/Oct/21
$${f}\left({x}\right)=\mid\mathrm{1}+{x}\mid+\mathrm{3}\mid\mathrm{2}+{x}\mid \\ $$$${f}'\left({x}\right)={sgn}\left(\mathrm{1}+{x}\right)+\mathrm{3}{sgn}\left(\mathrm{2}+{x}\right)\neq\mathrm{0}\:\forall{x}\in\mathbb{R}−\left\{−\mathrm{1},−\mathrm{2}\right\} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{3} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{1} \\ $$$${li}\underset{{x}\rightarrow\pm\infty} {{m}}\:{f}\left({x}\right)=\infty \\ $$$${min}\:{f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/Oct/21
$${thanks}\:{alot} \\ $$