# Derivative Problem

• Aug 15th 2009, 02:18 PM
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 $f'(x)$.
• Aug 15th 2009, 02:36 PM
NonCommAlg
Quote:

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 $f(x)=\frac{x^3}{3} + x.$