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  • Extreme values of implicit function

    Aruwin

    Aruwin

    @aruwin-WQ7eeu
    Updated: Oct 24, 2024
    Views: 1.7K
    Please check if my solution is correct or not.
    Find the extreme values of the following implicit function:
    2xy^3 + y - x^2 = 0
    My attempt:
    2xy³ + y - x² = 0
    2y³ + 6xy²(dy / dx) + dy / dx - 2x = 0
    6xy²(dy / dx) + dy / dx = 2x - 2y³
    (6xy² + 1)(dy / dx) = 2x - 2y³
    dy / dx = (x - y³) / (6xy² + 1)

    dy / dx = 0
    (x - y³) / (6xy² + 1) = 0
    x - y³ = 0
    y³ = x

    2y⁶ + y - y⁶ = 0
    y⁶ + y = 0
    y(y⁵ + 1) = 0
    y = 0 OR y⁵ + 1 = 0
    y = 0 OR y⁵ = -1
    y = 0 OR y = -1

    hence the extreme points:
    When y = 0, x = 0
    When y = -1, x = -1
    The extreme points are (-1, -1) and (0, 0).
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