Is $x=0$ a saddle point of the potential $V(x)=\frac{k}{4}x^4$?

Question

I'm a physics student. I came across a potential $V(x)$ of the form $$V(x)=\frac{k}{4}x^4$$ with $k>0$. For this potential $V^\prime(0)=0$ and also$V^{\prime\prime}(0)=0$ implies that the point $x=0$ is neither a maximum nor a minimum but a saddle point. But intuitively, if we plot the function $V(x)$, there is a plateau around $x=0$ which strongly suggests that it is a minimum and doesn't quite look like a saddle point.

Does my intuition fail or is $x=0$ really a saddle point?

Answer 1

The function $V$ in fact attains a global minimum at $x=0$. To see this, consider that $V(x)=\frac{k}{4} x^4\ge 0$ because $x^4=(x^2)^2\ge 0$ and $\frac{k}{4}>0$. Further, elementary algebra shows that $x^4=0$ has only one solution: namely $x=0$.

The second derivative test is inconclusive in the case that $f''(x_0)=0$. Such a point $x_0$ could be a minimum point, a maximum point, or neither – i.e. a saddle point.

Answer 2

since we have $$V^{(iv)}(x)=6k>0$$ and the order of the derivative is even we have there a Minimum see about Theorem of higher derivatives