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March 30, 2017

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Use analytical methods to find the exact global maximum and minimum values of the function f(x)=8x-ln(4x) for x >0. If a global maximum or minimum does not exist, enter the word NONE.

For the global maximum at x=none, But for the Global minimum at x=?

Is the Global minimum 0 or nonw? Any help would be greatly appreciated.

  • Calculus - ,

    I always encouraged my students to make a rough sketch of the graph.
    Pick any 5 or 6 positive x's, then find their y values.
    Pick an x close to zero (e.g. x = .00001) and see what y you get. This gives you an indication of the graph does close to zero. (remember x cannot be zero for this function)

    In general, the max/min is found by setting the derivative equal to zero, solving for the variable, and then subbing that back into the original equation.
    so
    f(x) = 8x - ln(4x)
    f'(x) = 8 - 1/x
    set this equal to zero, you will get x = 1/8

    I will let you find the minimum.

  • Calculus - ,

    y = 8 x - ln (4x)
    y' = 8 -4/x
    when is that zero?
    4/x = 8
    x = 1/2
    is that a maximum or a minimum?
    y" = 0 +4/x^2
    for x>0, y" is positive so it is a minimum
    so minimum at x = 1/2
    what is the value of the function at that minimum?
    y = 8(1/2) - ln 2
    4 - ln 2 = about 3.3

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