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Find the function f, given that the slope of the tangent line at any point (x,f(x)) is f '(x) and that the graph of f passes through the given point.
f '(x)=6(2x-7)^5 at (4, 3/2)

  • Math -

    f=INT f= (2x-7)^6 + C

    now find C

    3/2 = (2*4-7)^6 + C

    C=1/2

  • Math -

    if dy/dx = 6(2x-7)^5

    y = (6/6) (2x-7)^6 (1/2) + c
    y = (1/2)(2x-7)^6 + c
    but (4 , 3/2) lies on it, so

    3/2 = (1/2)(1)^6 + c
    3/2 = 1/2 + c
    c = 1

    f(x) = (1/2)(2x-7)^6 + 1

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