# Derivative Problem

• Aug 15th 2009, 02:18 PM
Derivative Problem
Suppose $\displaystyle f$ is a function that satisifies the equation $\displaystyle f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $\displaystyle x,y$. Suppose that $\displaystyle \lim_{x->0}\frac{f(x)}{x}=1$. Find $\displaystyle f(0),f'(0),f'(x)$.

I star by writing $\displaystyle x+y=0,x=-y, y=-x$

$\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
$\displaystyle =f(x)+f(-x)$

So now I have:

$\displaystyle f(x)=f(0)-f(-x)$

$\displaystyle f(0)=f(0)-f(0)=0$

For $\displaystyle f'(0)$:

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

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

So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.
• Aug 15th 2009, 02:36 PM
NonCommAlg
Quote:

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

I star by writing $\displaystyle x+y=0,x=-y, y=-x$

$\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
$\displaystyle =f(x)+f(-x)$

So now I have:

$\displaystyle f(x)=f(0)-f(-x)$

$\displaystyle f(0)=f(0)-f(0)=0$

For $\displaystyle f'(0)$:

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

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

So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.

to get $\displaystyle f(0) = 0$ it's easier to put $\displaystyle y = 0.$ to find $\displaystyle f'(x)$: $\displaystyle f'(x)=\lim_{y \to 0} \frac{f(x+y)-f(x)}{y}=\lim_{y\to0} \left(\frac{f(y)}{y} + x^2 + xy \right)=x^2 + 1.$ this also gives you $\displaystyle f(x)=\frac{x^3}{3} + x.$