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Calculus

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Suppose that f(x) = (5x + -2)^(1/2)

Find an equation for the tangent line to the graph of f(x) at x=1.

Tangent line: y=?

  • Calculus -

    y = (5x-2)^.5 ??? I guess from what you wrote
    dy/dx = .5 (5) (5x-2)^-.5
    at x = 1 :
    dy/dx = 2.5 / sqrt(3)
    that is our slope
    now at x = 1, y = sqrt(5-2) = sqrt 3
    y = (2.5/sqrt 3) x + b
    sqrt 3 = (2.5/sqrt 3)(1) + b

    b = sqrt 3 - (2.5/sqrt 3)
    so
    y = (2.5/sqrt 3) x + sqrt 3 -2.5/sqrt 3

    y sqrt 3 = 2.5 x +3 -2.5

    y sqrt 3 = 2.5 x + 1/2
    (2 sqrt 3) y = 5 x + 1

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