Total energy E_T =2 J It is fixed. For maximum speed. kinetic energy is maximum The potential energy should therefore be minimum

$\because V\left(x\right)=\frac{x^4}{4}-\frac{x^2}{2}$  

or $\frac{dV}{dx}=\frac{4x^3}{4}-\frac{2x}{2} =x(x^2-1)$

For V to be minimum,$\frac{dV}{dx}=0$  

$\therefore x(x^2-1)=0,\mathrm{or }x=0,\pm 1$  

At x=0, V(x)=0

At $x=\pm 1,V\left(x\right)=-\frac{1}{4}\text{J}$  

$\therefore (\text{Kinetic energy}{)}_{\text{max}}=E_T-V_{\text{min}}$  

or $\therefore (\text{Kinetic energy}{)}_{\text{max}}=\left(\frac{1}{4}\right)=\frac{9}{4} \text{J}$ 

or $\frac{1}{2}m{v}_m^2=\frac{9}{4}$ 

$v_{\text{m}}^2=\frac{9\times 2}{\text{m}\times 4}$

or $v_m^2=\frac{9\times 2}{m\times 4}=\frac{9\times 2}{1\times 4}$

$\therefore v_m=\frac{3}{\sqrt{2}} \mathrm{m} {\mathrm{s}}^{-1}$