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Calculus

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Find the tangent line approximation for sqrt(3+x) near x=0.

  • Calculus -

    when x = 0, y = sqrt 3

    y = (x+3)^.5
    so
    dy/dx = .5 (x+3)^-.5
    slope of tangent line at x = 0 is thus
    .5/sqrt 3
    y = m x + b
    sqrt 3 = (.5/sqrt 3)(0) + b
    b = sqrt 3

    y = (.5/sqrt 3) x + sqrt 3

  • Calculus -

    Thank you

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