Calculus

Find a function of the form f(x)=a+bsqrtx that is tangent to the line 2y-3x=5 at the point (1, 4)

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  1. from 2y-3x=5 , we know that the slope of the tangent must be 3/2

    f(x) = a + b√x = a +b(x)^(1/2)
    f ' (x) = (1/2)b x^(-1/2) or b/2√x

    so b/2√x = 3/2
    2b = 6√x
    b = 3√x
    also (1,4) lies on f(x) = a + bx
    4 = a + b
    since b = 3√x = 3√1 = 3
    then a = 1

    f(x) = 1 + 3√x is our function

    check:
    f ' (x) = (3/2)x^(-1/2)
    at (1,4) , slope = (3/2)(1/√1) = 3/2
    equation of tangent:
    y-4 = (3/2)(x-1)
    2y - 8 = 3x - 3
    2y - 3x = 5, as given

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