# 一定要分清拐点和极值点：极值点是一阶导等于零的点，拐点是二阶导等于零的点

## 二、解析

### 错误的解法（误认为拐点是一阶导等于零的点）

$$y=x^{3}+a x^{2}+b x+14 \Rightarrow$$

$$y^{\prime}=3 x^{2}+2 a x+b \Rightarrow$$

$$\left\{\begin{array}{l}y^{\prime}(1)=0 \Rightarrow 3+2 a+b=0 \\ y(1)=3 \Rightarrow 1+a+b+14=3 \end{array}\right. \Rightarrow$$

$$\left\{\begin{array}{l}3+2 a+b=0 \\ a+b=3-15 \end{array}\right. \quad \Rightarrow$$

$$\left\{\begin{array}{l}3+2 a+b=0 \\ a=-b-12\end{array}\right. \Rightarrow$$

$$3-2(b+12)+b=0 \Rightarrow$$

$$3-2 b-24+b=0 \Rightarrow b=-21 \Rightarrow$$

$$a=9 \Rightarrow$$

$$\left\{\begin{array}{l}a=9 \\ b=-21\end{array}\right.$$

### 正确的解法

$$y=x^{3}+a x^{2}+b x+14 \Rightarrow$$

$$y^{\prime}=3 x^{2}+2 a x+b \Rightarrow$$

$$y^{\prime \prime}=6 x+2 a \Rightarrow$$

$$y^{\prime \prime}(1)=0 \Rightarrow 6+2 a=0 \Rightarrow \textcolor{springgreen}{ a=-3 }$$

$$y=x^{3}-3 x^{2}+b x+14 \Rightarrow$$

$$y(1)=3 \Rightarrow \textcolor{orangered}{ 1-3+b+14=3 } \Rightarrow$$

$$b+15 – 3=3 \Rightarrow$$

$$b=3-12 \Rightarrow \textcolor{springgreen}{ b = -9 }$$

$1 \textcolor{orangered}{ -3 } +b+14$ $=$ $\textcolor{orangered}{ 3 }$ $\nRightarrow$ $1 + 0+b+14=0$