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Math Help - Derivative Problem

  1. #1
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    Derivative Problem

    Suppose f is a function that satisifies the equation f(x+y)=f(x)+f(y)+yx^2+xy^2 for all real x,y. Suppose that \lim_{x->0}\frac{f(x)}{x}=1. Find f(0),f'(0),f'(x).

    I star by writing x+y=0,x=-y, y=-x

    f(0)=f(x)+f(-x)+x^3-x^3
    =f(x)+f(-x)

    So now I have:

    f(x)=f(0)-f(-x)

    f(0)=f(0)-f(0)=0

    For f'(0):

    f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}

    =\lim_{x->0}\frac{f(x)}{x}=1

    So I got f'(0)=1,f(0)=0. I can't seem to get started on finding f'(x).
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    Suppose f is a function that satisifies the equation f(x+y)=f(x)+f(y)+yx^2+xy^2 for all real x,y. Suppose that \lim_{x->0}\frac{f(x)}{x}=1. Find f(0),f'(0),f'(x).

    I star by writing x+y=0,x=-y, y=-x

    f(0)=f(x)+f(-x)+x^3-x^3
    =f(x)+f(-x)

    So now I have:

    f(x)=f(0)-f(-x)

    f(0)=f(0)-f(0)=0

    For f'(0):

    f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}

    =\lim_{x->0}\frac{f(x)}{x}=1

    So I got f'(0)=1,f(0)=0. I can't seem to get started on finding f'(x).
    to get f(0) = 0 it's easier to put y = 0. to find f'(x): f'(x)=\lim_{y \to 0} \frac{f(x+y)-f(x)}{y}=\lim_{y\to0} \left(\frac{f(y)}{y} + x^2 + xy \right)=x^2 + 1. this also gives you f(x)=\frac{x^3}{3} + x.
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