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Calculus

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Find k such that the line is tangent to the graph of the function. Function: f(x)=kã(x)
Line: y=x+4

  • Calculus -

    I will assume your equation is
    f(x) = k√x = k x^(1/2)

    then f'(x) = (k/2)x^(-1/2)
    but the slope of y = x+4 is 1

    so (k/2)x^(-1/2) = 1
    k/(2√x) = 1
    k = 2√x

    then y = 2√x √x = 2x

    sub back into y = x+4
    2x = x+4
    x = 4

    so k= 2√4 = 4

  • Calculus -

    y=2x²√ 2-x

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