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Homework Help: Minimum value of the expression

  1. Jul 17, 2014 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    Problem is posted as image


    2. Relevant equations



    3. The attempt at a solution
    Hello,

    I am having some confusion over what is meant by 'type in the boxes the minimum value of the expression'. Does that mean take the derivative of the function? Or does that mean the value at which the function is a minimum? That would be setting them all to zero

    a) ##\underset \min{x} \hspace{0.05 in} x = ##

    [itex] f(x) = x [/itex]
    [itex] f'(x) = 1 [/itex]
    [itex] f'(x) = 0 [/itex] at the minimum
    [itex] 1 \neq 0 [/itex]
    [itex] \bar{x} = -\infty [/itex]

    b) ## \underset \min{x}\hspace{0.05 in}2x^2 = ##

    [itex] f(x) = 2x^2 [/itex]
    [itex] f'(x) = 4x [/itex]
    [itex] f'(x) = 0 [/itex] at the minimum
    [itex] 4x = 0 [/itex]
    [itex] \bar{x} = 0 [/itex]


    c) ##\underset \min{x} \hspace{0.05 in}x + 2x^2 = ##

    [itex] f(x) = x + 2x^2 [/itex]
    [itex] f'(x) = 4x + 1 [/itex]
    [itex] f'(x) = 0 [/itex] at the minimum
    [itex] 4x = -1 [/itex]
    [itex] \bar{x} = -0.25 [/itex]

    d) ##\underset \min{x}\hspace{0.05 in} 5 - x + 2x^2 = ##

    [itex] f(x) = x + 2x^2 [/itex]
    [itex] f'(x) = 4x - 1 [/itex]
    [itex] f'(x) = 0 [/itex] at the minimum
    [itex] 4x = 1 [/itex]
    [itex] \bar{x} = 0.25 [/itex]
     

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    Last edited: Jul 17, 2014
  2. jcsd
  3. Jul 17, 2014 #2

    Mentallic

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    Homework Helper

    It means to write the y value of the function that occurs at the minimum x value.
    For the parabola

    [tex]x+2x^2[/tex]

    the min occurs at [itex]x=-0.25[/itex] and hence the answer would be [tex]f(-0.25)=-0.25+2(-0.25)^2=-0.125[/tex]
     
  4. Jul 17, 2014 #3

    Maylis

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    Gold Member

    Thanks, got it.
     
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