Question Number 114768 by ZiYangLee last updated on 21/Sep/20

Solve for ∣x−2∣+∣1−x∣=4

Answered by bobhans last updated on 21/Sep/20

(1) x<1 ⇒ 2−x+1−x=4 3−2x = 4 ; x=−(1/2) (2) 1≤x<2⇒2−x+x−1=4 1=4 ←no solution (3)x≥2 ⇒x−2+x−1=4 2x = 7; x=(7/2) ∴ x = −(1/2) or x = (7/2)

Answered by mathmax by abdo last updated on 22/Sep/20

⇒f(x)=∣x−2∣+∣1−x∣−4 =0 x −∞ 1 2 +∞ ∣x−2∣ −x+2 −x+2 x−2 ∣x−1∣ −x+1 x−1 x−1 f(x) −2x−1 −3 2x−7 ⇒f(x) = { ((−2x−1 if x≤1)),((−3 if 1≤x≤2 )) :} { 2x−7 if x≥2 so x≤1 we get f(x)=0 ⇒−2x−1 =0 ⇒x =−(1/2) 1≤x≤2 f(x)=0 ⇒−3 =0 imposdible x≥2 f(x)=0 ⇒2x−7 =0 ⇒x =(7/2)