Show f(x) Is a Minimum

Discussion in 'Other Pre-University Math' started by nycmathguy, Nov 14, 2021.

  1. nycmathguy

    nycmathguy

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    David Cohen

    How is this done? Seeking hints.

    20211113_195913.jpg
     
    nycmathguy, Nov 14, 2021
    #1
  2. nycmathguy

    MathLover1

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    f(x)=(x-a)^2+(x-b)^2+(x-c)^2

    show that f(x) will be a minimum when x is the avarage of a,b, and c

    derivate

    f(x)=2(x-a)+2(x-b)+2(x-c)..........equal to 0
    0=2x-2a+2x-2b+2x-2c
    2x+2x+2x-2a-2b-2c=0
    6x-2(a+b+c)=0
    6x=2(a+b+c)
    x=2(a+b+c)/6
    x=(a+b+c)/3 which the avarage of a,b, and c
     
    MathLover1, Nov 16, 2021
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Excellent. Nice job.
     
    nycmathguy, Nov 16, 2021
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