# Function

• Feb 18th 2009, 03:47 AM
Ortega
Function
Detrmine f(x)=?? if:
f(x)+f[x]+f{x}=x
• Feb 18th 2009, 07:23 AM
red_dog
If $\displaystyle x=0\Rightarrow 3f(0)=0\Rightarrow f(0)=0$

If $\displaystyle x\in\mathbf{Z}\Rightarrow [x]=x, \ \{x\}=0$

Then $\displaystyle f(x)+f([x])+f(\{x\})=x\Rightarrow f(x)+f(x)+f(0)=x\Rightarrow 2f(x)=x\Rightarrow$
$\displaystyle \Rightarrow f(x)=\frac{x}{2}, \ \forall x\in\mathbf{Z}$

If $\displaystyle x\in[0,1)\Rightarrow [x]=0, \ \{x\}=x$
Then $\displaystyle f(x)+f([x])+f(\{x\})=x\Rightarrow f(x)+f(0)+f(x)=x\Rightarrow$

$\displaystyle \Rightarrow 2f(x)=x\Rightarrow f(x)=\frac{x}{2}, \ \forall x\in[0,1)$

Now, let $\displaystyle x\in\mathbf{R}\Rightarrow x=[x]+\{x\}, \ [x]\in\mathbf{Z}, \ \{x\}\in[0,1)$

Then $\displaystyle f(x)+f([x])+f(\{x\})=x\Rightarrow f(x)+\frac{[x]}{2}+\frac{\{x\}}{2}=x\Rightarrow$

$\displaystyle \Rightarrow f(x)=x-\frac{[x]+\{x\}}{2}=x-\frac{x}{2}=\frac{x}{2}$

So, $\displaystyle f(x)=\frac{x}{2}, \ \forall x\in\mathbf{R}$