$f$ is defined on $[a,b]$,$a,b\in\mathbf{R},a<b$.$x_0\in [a,b]$.If $f$ is differentiable ,and second differentiable on $[a,b]$,then

\begin{equation}

\label{eq:29.12.51}

f'(x_0)=0,f''(x_0)>0\Rightarrow x_0 ~\mbox{is a local minimum}

\end{equation}

Proof:

\begin{equation}

\label{eq:29.12.52}

f''(x_0)=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{f'(x_0+\Delta

x)-f'(x_0)}{\Delta x}>0

\end{equation}

So,when $\Delta x>0$ and $\Delta x$ is small enough,$f'(x_0+\Delta x)-f'(x_0)>0$,so $f'(x_0+\Delta x)>0$.                                                                                                    

When $\Delta x<0$ and $|\Delta x|$ is small enough,$f'(x_0+\Delta x)-f'(x_0)<0$,so $f'(x_0+\Delta x)<0$.So $x_0$ is a local minimum(why?According to the differential mean value theorem.)

 

Remark 1:Similarly,\begin{equation}

\label{eq:29.13.58}

f'(x_0)=0,f''(x_0)<0\Rightarrow x_0 ~\mbox{is a local maximum}

\end{equation}