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Use a graphing utility to approximate any relative minimum or maximum values of the function g(x)=x sqrt4-x Not getting this one, either. Thanks.

  • typo? -

    do you mean what you wrote or
    x sqrt (4-x) ????

  • Pre-calculus -

    That's because there is no maximum/minimum

    g(x) = √(4-x) = (4-x)^(1/2)
    g '(x) = (1/2)(4-x)^(-1/2) (-1)
    = -1/(2√(4-x)
    This can never be zero

  • Pre-calculus -

    if the latter
    y = x (4-x)^.5
    dy/dx = x (.5)(4-x)^-.5)(-1) + (4-x)^.5
    when is that zero
    0 = -.5x /(4-x)^.5 + (4-x)^.5
    0 = -.5 x + (4-x)
    4 = 1.5 x
    x = 2.67 for max or min
    You should find a max or min around there with your grapher
    You can find out with the second derivative but you should see it on your graph anyway
    I will find a grapher online
    for function
    x * sqrt(4-x)

  • Pre-calculus -

    Here you go

    put in

    x * sqrt(4-x)

    and hit enter. You will see the max at 2.67

  • oops - Pre-calculus -

    didn't see that x in front of √

    so by the product rule
    I got
    g'(x) = (1/2)(3x-8)(4-x)^(-1/2)
    which when we set that to zero
    x = 8/3

    your graphing calculator should show a max point near x = 8/3 , (near about x=3)

  • Pre-calculus -

    Why is there a .5 in your problem? And yes, it's x sqrt(4-x)

  • ^.5??? -

    Is ^.5 supposed to be square root? Never heard it put that way before.

  • Pre-calculus -

    easy way to write square root - to the 1/2 power

  • Pre-calculus -

    especially when doing derivatives
    d/dx x^(1/2) = (1/2) x^-(1/2)
    d/dx x^.5 = .5 x^-.5

  • Pre-calculus -


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