calculus

Find the critical numbers x1 and x2 of y=(9x^2+3)/4x.

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  1. y = (9x^2 + 3)/x
    dy/dx = (x(18x) - (9x^2 + 3)/x^2
    = (18x^2 - 9x^2 - 3)/x^2
    = 0 for critical values

    9x^2 = 3
    x^2 = 1/3
    x = ± 1/√3

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