calculus

Find the equation, in slope intercept form , of the line tangent to the graph of the square root of (x^3+7x+1), when x=3

asked by Patty
  1. when x = 3
    y = (3^3 + 21 + 1)^.5 = sqrt(49) = 7 (or -7 but we will pretend we did not know that)

    dy/dx = (1/2)(x^3+7x+1)^(-1/2) (3 x^2)
    at x =3 this is
    (1/2)(1/7)(27) = 27/14 = slope of line

    y = (27/14) x + b
    7 = 27(3)/14 + b
    solve for b

    posted by Damon

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