Let f(x) be a real function not identically zero in Z, such that $f\left(x+y^{2n+1}\right)=f\left(x\right)+{\left\{f\left(y\right)\right\}}^{2n+1},n\in N$  and $x,y\in R$ . If $f^1\left(0\right)\ge 0$  then $f^1\left(6\right)$ is equal to

$f\left(0\right)=0$ 

$f\left(1\right)=f\left(0\right)+{\left\{f\left(1\right)\right\}}^{2n+1}$

$f\left(1\right)\left[f{\left(1\right)}^{2n}-1\right]=0$

$f\left(1\right)=0or1or=-1$

Now, $f^1\left(0\right)=\underset{x\to 0}{lt}\frac{f\left(x\right)-f\left(0\right)}{n}=\underset{x\to 0}{lt}\frac{f\left(x\right)}{x}\ge 0$

$\Rightarrow f\left(x\right)\ge 0forx>0\therefore f\left(1\right)\ne -1$

Also $f\left(1\right)=0$is not true $\left(\because f\left(2\right)=f\left(3\right)=\mathrm{.......0}\right)$

$\Rightarrow f\left(1\right)=1$

 $\therefore f\left(x\right)$ is a liner function passing  through$\left(0,0\right)$

 $\Rightarrow f\left(x\right)=x$        $\Rightarrow f^1\left(6\right)=1$