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Calculus

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Find the critical numbers of y=x(4-x)^1/2.
The answer is x=8/3 but I got x=7/2.

  • Calculus -

    y = x√(4-x)
    y' = 1√(4-x) + x * 1/2 * 1/√(4-x) * -1
    = √(4-x) - x/(2√(4-x))
    = (2(4-x) - x)/(2√(4-x))
    = (8-3x)/(2√(4-x))

    assuming x≠4, we just have to find where the numerator is zero.

    x = 8/3

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