Ngô Quốc Anh

April 1, 2008

Find all derivable functions

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 22:25

Find all derivable functions $\mathbb R\to \mathbb R$, with the property

$f(x+y+z)+3xyz=f(x)+f(y)+f(z)+3(x+y+z)(xy+yz+zx)-2$

for every $x,y,z\in \mathbb R$.

Solution. $'_x$ with respect to $x$ to both sides

$f'\left( {x + y + z} \right) - 3\left( {x + y + z} \right)^2 = f'\left( x \right) - 3x^2$.

Thus

$f'\left( x \right) - 3x^2$

is a constant.

Note. Actually with only ”$f$ is derivable in one point” you obtain the same result

$x=y=z=0 \Longrightarrow f(0)=1$.

Suppose that $f$ is derivable in $t_0$

$(x=t_0)\& (z=0) \Longrightarrow \frac {f(y)-1}y = \frac {f(t_0+y)-f(t_0)}y-3t_0(t_0+y)$,

so $f$ is derivable in $0$. But

$\frac {f(x+y)-f(x)}y = 3x(x+y)+\frac {f(y)-1}y$

shows that $f(x)$ is derivable for all $x\in \mathbb R$: $f'(x) = 3x^2 + f'(0)$ and so on.