Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 17872 by Tinkutara last updated on 11/Jul/17

Solve :  ∣x − 1∣ + ∣x∣ + ∣x + 1∣ = x + 2

$$\mathrm{Solve}\:: \\ $$$$\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\mid\:+\:\mid{x}\:+\:\mathrm{1}\mid\:=\:{x}\:+\:\mathrm{2} \\ $$

Answered by ajfour last updated on 11/Jul/17

   x∈[0, 1]

$$\:\:\:\mathrm{x}\in\left[\mathrm{0},\:\mathrm{1}\right] \\ $$

Answered by ajfour last updated on 12/Jul/17

let f(x)=∣x−1∣+∣x∣+∣x+1∣    f(x)= { ((−3x  ;      x≤−1)),((2−x  ;  −1≤x≤0)),((2+x  ;      0≤x≤1)),((   3x   ;      1≤x)) :}

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{x}\mid+\mid\mathrm{x}+\mathrm{1}\mid \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{−\mathrm{3x}\:\:;\:\:\:\:\:\:\mathrm{x}\leqslant−\mathrm{1}}\\{\mathrm{2}−\mathrm{x}\:\:;\:\:−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{0}}\\{\mathrm{2}+\mathrm{x}\:\:;\:\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}}\\{\:\:\:\mathrm{3x}\:\:\:;\:\:\:\:\:\:\mathrm{1}\leqslant\mathrm{x}}\end{cases} \\ $$

Commented by ajfour last updated on 12/Jul/17

see Q.17895  for graph.

$$\mathrm{see}\:\mathrm{Q}.\mathrm{17895}\:\:\mathrm{for}\:\mathrm{graph}. \\ $$

Commented by Tinkutara last updated on 12/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com