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Question Number 119052 by shahria14 last updated on 21/Oct/20

Answered by 1549442205PVT last updated on 22/Oct/20

∣x^2 −1∣≤3⇔−3≤x^2 −1≤3⇔−2≤x^2 ≤4 ⇔x^2 ≤4⇔∣x∣≤2⇔−2≤x≤2

∣x2−1∣⩽3⇔−3⩽x2−1⩽3⇔−2⩽x2⩽4⇔x2⩽4⇔∣x∣⩽2⇔−2⩽x⩽2

Commented by MJS_new last updated on 22/Oct/20

ok but now solve this: ∣x^2 −5∣≤3

okbutnowsolvethis:∣x2−5∣⩽3

Commented by bemath last updated on 23/Oct/20

min ∣x^2 −5∣ = −5 ∧ ∣−5∣ > 3 ⇒ −3 ≤ x^2 −5≤3 ⇒ 2 ≤ x^2 ≤ 8 → { ((x^2 ≥2 ⇒ x≤−(√2) ∪ x ≥(√2))),((x^2 ≤8⇒−2(√2) ≤ x ≤ 2(√2))) :} the solution is −2(√2) ≤x≤−(√2) ∪ (√2) ≤x≤2(√2)

min∣x2−5∣=−5∧∣−5∣>3⇒−3⩽x2−5⩽3⇒2⩽x2⩽8→{x2⩾2⇒x⩽−2∪x⩾2x2⩽8⇒−22⩽x⩽22thesolutionis−22⩽x⩽−2∪2⩽x⩽22

Answered by MJS_new last updated on 22/Oct/20

since min (x^2 −1) = −1 and ∣−1∣≤3 ⇒ it′s enough to calculate x^2 −1≤3 ⇔ x^2 ≤4 ⇔ −2≤x≤2

sincemin(x2−1)=−1and∣−1∣⩽3⇒it′senoughtocalculatex2−1⩽3⇔x2⩽4⇔−2⩽x⩽2

Commented by bemath last updated on 22/Oct/20

typo min (x^2 −1) =−1

typomin(x2−1)=−1

Commented by MJS_new last updated on 22/Oct/20

yes... the display of my smartphone is too narrow, so sometimes I hit 2 keys at the same time...

yes...thedisplayofmysmartphoneistoonarrow,sosometimesIhit2keysatthesametime...

Answered by Bird last updated on 21/Oct/20

∣x^2 −1∣≤3 ⇒∣x^2 −1∣−3≤0 ⇒f(x)≤0 sith f(x)=∣x^2 −1∣−3 x −∞ −1 1 +∞ ∣x^2 −1∣ x^2 −1 0 1−x^2 0 x^2 −1 f(x) x^2 −4 −2−x^2 x^2 −4 if x≤−1 f(x)≤0 ⇒x^2 −4 ≤0 ⇒ −2≤x≤2 ⇒S_1 =[−2,−1] if x∈[−1,1] f(x)≤0 ⇒−2−x^2 ≤0 ⇒ 2+x^2 ≥0 ⇒S_2 =[−1,1] if x≥1 f(x)≤0 ⇒x^2 −4≤0 ⇒ −2≤x≤2 ⇒S_3 =[1,2] ⇒ S=∪S_i =[−2,2]

∣x2−1∣⩽3⇒∣x2−1∣−3⩽0⇒f(x)⩽0sithf(x)=∣x2−1∣−3x−∞−11+∞∣x2−1∣x2−101−x20x2−1f(x)x2−4−2−x2x2−4ifx⩽−1f(x)⩽0⇒x2−4⩽0⇒−2⩽x⩽2⇒S1=[−2,−1]ifx∈[−1,1]f(x)⩽0⇒−2−x2⩽0⇒2+x2⩾0⇒S2=[−1,1]ifx⩾1f(x)⩽0⇒x2−4⩽0⇒−2⩽x⩽2⇒S3=[1,2]⇒S=∪Si=[−2,2]

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