9.
有一段“三段论”推理是这样的:对于可导函数\(f\left(x\right) \),如果\({f}^{{{{"}}}}\left({x}_{0}\right)=0 \),那么\(x={x}_{0} \)是函数\(f\left(x\right) \)的极值点,因为函数\({{z}_{2}}-{{z}_{1}}\)在\(x=0\)处的导数值\({f}^{{{{"}}}}\left(0\right)=0 \),所以,\(x=0\)是函数\(f\left(x\right)={x}^{3} \)的极值点\(.\)以上推理中( )