Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 157604 by mnjuly1970 last updated on 25/Oct/21

          Question    find the” minimum” value of:    f (x):= ∣1+x∣+∣ 2+x∣ + ∣4 +2x∣

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Q}{uestion}\: \\ $$$$\:{find}\:{the}''\:{minimum}''\:{value}\:{of}: \\ $$$$ \\ $$$${f}\:\left({x}\right):=\:\mid\mathrm{1}+{x}\mid+\mid\:\mathrm{2}+{x}\mid\:+\:\mid\mathrm{4}\:+\mathrm{2}{x}\mid \\ $$$$ \\ $$

Commented by mr W last updated on 25/Oct/21

i think the easiest way is to check all  kink positions:  x=−1, −2  f(−1)=0+1+2=3  f(−2)=1+0+0=1  ⇒f(x)_(min) =1

$${i}\:{think}\:{the}\:{easiest}\:{way}\:{is}\:{to}\:{check}\:{all} \\ $$$${kink}\:{positions}: \\ $$$${x}=−\mathrm{1},\:−\mathrm{2} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{0}+\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{1}+\mathrm{0}+\mathrm{0}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)_{{min}} =\mathrm{1} \\ $$

Commented by mr W last updated on 25/Oct/21

in this case we can also do like this:  f(x)=∣1+x∣+3∣1+1+x∣=∣1+x∣+3±3∣1+x∣  =3+ { ((4∣1+x∣)),((−2∣1+x∣)) :}  f(x)_(min) =3−2=1 at x=−1

$${in}\:{this}\:{case}\:{we}\:{can}\:{also}\:{do}\:{like}\:{this}: \\ $$$${f}\left({x}\right)=\mid\mathrm{1}+{x}\mid+\mathrm{3}\mid\mathrm{1}+\mathrm{1}+{x}\mid=\mid\mathrm{1}+{x}\mid+\mathrm{3}\pm\mathrm{3}\mid\mathrm{1}+{x}\mid \\ $$$$=\mathrm{3}+\begin{cases}{\mathrm{4}\mid\mathrm{1}+{x}\mid}\\{−\mathrm{2}\mid\mathrm{1}+{x}\mid}\end{cases} \\ $$$${f}\left({x}\right)_{{min}} =\mathrm{3}−\mathrm{2}=\mathrm{1}\:{at}\:{x}=−\mathrm{1} \\ $$

Commented by mnjuly1970 last updated on 25/Oct/21

grateful sir W

$${grateful}\:{sir}\:{W} \\ $$

Answered by TheSupreme last updated on 25/Oct/21

f(x)=∣1+x∣+3∣2+x∣  f′(x)=sgn(1+x)+3sgn(2+x)≠0 ∀x∈R−{−1,−2}  f(−1)=3  f(−2)=1  lim_(x→±∞)  f(x)=∞  min f(x)=1

$${f}\left({x}\right)=\mid\mathrm{1}+{x}\mid+\mathrm{3}\mid\mathrm{2}+{x}\mid \\ $$$${f}'\left({x}\right)={sgn}\left(\mathrm{1}+{x}\right)+\mathrm{3}{sgn}\left(\mathrm{2}+{x}\right)\neq\mathrm{0}\:\forall{x}\in\mathbb{R}−\left\{−\mathrm{1},−\mathrm{2}\right\} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{3} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{1} \\ $$$${li}\underset{{x}\rightarrow\pm\infty} {{m}}\:{f}\left({x}\right)=\infty \\ $$$${min}\:{f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 25/Oct/21

thanks alot

$${thanks}\:{alot} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com