If  a differentiable function f(x) satisfies the equation $f\left(x-2\right)+f\left(x\right)+f\left(x+2\right)=0,\forall x\in R$ and$f^{\text{'}}\left(0\right)=4$  , then select incorrect option

 $\therefore f\left(x-2\right)+f\left(x\right)+f\left(x+2\right)=0\mathrm{.......}\left(i\right)$
Replace x by$x+2$  , we get
$f\left(x\right)+f\left(x+2\right)+f\left(x+4\right)=0\mathrm{........}\left(ii\right)$
From (i) and (ii)
$f\left(x-2\right)=f\left(x+4\right)$
Replace x by $x+2$ :  $f\left(x\right)=f\left(x+6\right)$
f(x) is periodic with period 6
and$f^{\text{'}}\left(x\right)$  is also periodic with period 6.