Suppose $f:[a,b]\to\mathbb{R}$ is a smooth ($C^\infty$) function which attains a strict minimum at some point $x_0\in (a,b)$. That is, $$ f(x_0) < f(x) \qquad\forall x\in[a,b]\setminus\{x_0\} $$ Suppose furthermore that $f^{\prime}(x_0)=0$ and $f^{\prime\prime}(x_0) >0$.
Does there exists a neighbourhood $V$ of $x_0$ such that $$f(x)\leq f(y)\qquad \forall x\in V,\, y\in[a,b]\setminus V$$
(I assume that by "neighborhood of $x_0$" you mean "interval containing $x_0$"; all the more general definitions of "neighborhood" would make this problem trivial.)
As $f''(x_0)>0$ and $f''$ is continuous, $f$ is strictly convex on some open interval $V_0$ containing $x_0$. Moreover, $[a, b] \backslash V_0$ is compact, and so $f$ attains some minimum value $y_1$ on $[a, b] \backslash V_0$.
Set $V=f^{-1}((-\infty,y_1))$. This contains $x_0$ because $x_0$ is a strict minimum. It is contained in $V_0$ by construction. But the convexity of $f$ on $V_0$ then implies that $V$ is an open interval: it is open because $f$ is continuous, and if $x_1,x_2 \in V$ then $f(x)<\max(f(x_1),f(x_2))$ for all $x \in (x_1,x_2)$. Finally, $f(x)<y_1$ for $x \in V$ and $f(x) \geq y_1$ for $x \not\in V$. So we are done.