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.

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