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Let-f-x-x-2-x-4-2x-6-for-2-x-8-The-sum-of-the-largest-and-smallest-values-of-f-x-is-




Question Number 110359 by Aina Samuel Temidayo last updated on 28/Aug/20
Let  f(x)=∣x−2∣+∣x−4∣−∣2x−6∣,  for 2≤x≤8. The sum of  the largest and  smallest values of f(x) is
Letf(x)=∣x2+x42x6,for2x8.Thesumofthelargestandsmallestvaluesoff(x)is
Answered by bobhans last updated on 28/Aug/20
∣x−2∣ =  { ((x−2 ; x ≥2)),((−x+2; x< 2)) :}  ∣x−4∣ =  { ((x−4 ; x ≥ 4)),((−x+4 ; x<4)) :}  ∣2x−6∣ =  { ((2x−6 ; x≥3)),((−2x+6 ; x<3)) :}  given 2≤ x≤ 8 →2≤x<3 ∪ 3≤x<4 ∪ 4≤x≤8  case(1) 2≤x<3→f(x)=x−2−(x−4)+2x−6                          f(x)=2x−4→ { ((f(2)=0)),((f(3)=2)) :}  case(2)3≤x<4→f(x)=x−2−(x−4)−(2x−6)                       f(x)=−2x+8 → { ((f(3)=2)),((f(4)=0)) :}  case(3)4≤x≤8→f(x)=x−2+x−4−(2x−6)                         f(x)=0   min value of f(x) = 0 & max value of  f(x) = 2  Σ(min, max) = 2
x2={x2;x2x+2;x<2x4={x4;x4x+4;x<42x6={2x6;x32x+6;x<3given2x82x<33x<44x8case(1)2x<3f(x)=x2(x4)+2x6f(x)=2x4{f(2)=0f(3)=2case(2)3x<4f(x)=x2(x4)(2x6)f(x)=2x+8{f(3)=2f(4)=0case(3)4x8f(x)=x2+x4(2x6)f(x)=0minvalueoff(x)=0&maxvalueoff(x)=2Σ(min,max)=2
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Wow. Thanks. Please help in the  remaining questions I posted.
Wow.Thanks.PleasehelpintheremainingquestionsIposted.
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Please how did you get this: given  2≤x≤8→ 2≤x<3 ∪ 3≤x<4 ∪ 4≤x≤8
Pleasehowdidyougetthis:given2x82x<33x<44x8

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