A particle free to move along the $\mathrm{X}\text{-axis }$has potential energy given by $\text{U(x)=k}\left[1-\exp \left(-{\mathrm{x}}^2\right)\right]$ for $-\mathrm{\infty }\le \mathrm{X}\le +\mathrm{\infty }$ where k is a positive constant of
appropriate dimensions. Then

Given that $\mathrm{U}\left(\mathrm{x}\right)=\mathrm{k}\left[1-{\mathrm{e}}^{-{\mathrm{x}}^2}\right]$
The graph of U(x) is shown in fig.   
 

$$\begin{gathered} \text{ Now, }F\left(x\right)=-\frac{dU\left(x\right)}{dx}=-2kx{e}^{-x^2} \\ Thus force is zero only at following three points : \\ x=0,x\to +\infty \text{ and }x\to -\infty \\ At any point away from origin (excluding \\ x\to +\infty \text{ and }x\to -\infty \text{ ) the particle is not even in } \\ equilibrium. \\ Further, at finite non-zero values of x, the forco is \\ directed towards origin. \\ Now F(x)=-2k\times e^{-x^2} \\ =-2kx\left[1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\dots \right] \\ 2k x if x<<1 \\ If displaced a little about origin, the particle will execute S.H.M. \\ Again, Total mechanical energy =K.E+P.E \\ \frac{k}{z}=K.E.+k\left[1-e^{-x^2}\right] \\ K.E.=k\left[-\frac{1}{2}+e^{-x^2}\right] \\ The graph of K.E. is also shom in fig. It is obvious from the gaph that K.E. is not the minimum at the origin. \end{gathered}$$