# Challenge problem.

• Feb 21st 2010, 05:24 AM
Ted
Challenge problem.
Hello.
I have this problem, I stucked.
I did not post it in the challenge sub-forum; because I do not know the solution.

Problem: Suppose $\displaystyle f$ is a function satisfies the equation $\displaystyle f(x+y)=f(x)+f(y)+x^2y+xy^2$ for all number $\displaystyle x$ and $\displaystyle y$. Suppose also that: $\displaystyle \lim_{x\to 0} \frac{f(x)}{x}=1$.
• (a) Find $\displaystyle f(0)$.
• (b) Find $\displaystyle f'(0)$.
• (c) Find $\displaystyle f'(x)$.
• Feb 21st 2010, 05:55 AM
skeeter
Quote:

Originally Posted by Ted
Hello.
I have this problem, I stucked.
I did not post it in the challenge sub-forum; because I do not know the solution.

Problem: Suppose $\displaystyle f$ is a function satisfies the equation $\displaystyle f(x+y)=f(x)+f(y)+x^2y+xy^2$ for all number $\displaystyle x$ and $\displaystyle y$. Suppose also that: $\displaystyle \lim_{x\to 0} \frac{f(x)}{x}=1$.
• (a) Find $\displaystyle f(0)$.
• (b) Find $\displaystyle f'(0)$.
• (c) Find $\displaystyle f'(x)$.

(a) $\displaystyle f(0+0) = f(0) + f(0) + 0^2 \cdot 0 + 0 \cdot 0^2$

$\displaystyle f(0) = 2f(0)$

$\displaystyle f(0) = 0$

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

$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x)}{x} = 1$

(c) $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x) + f(h) + x^2h + xh^2 - f(x)}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(h) + x^2h + xh^2}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \left[\frac{f(h)}{h} + \frac{h(x^2 + xh)}{h}\right]$

$\displaystyle f'(x) = \lim_{h \to 0} \left[\frac{f(h)}{h} + (x^2 + xh) \right]$

$\displaystyle f'(x) = 1 + x^2$