# 2023年考研数二第21题解析：泰勒公式、存在性的理解

## 二、解析

### 第 (1) 问

\textcolor{yellow}{ \begin{aligned} f ( x ) \\ \\ & = f ( 0 ) + f ^ { \prime } ( 0 ) x + \frac { f ^ { \prime \prime } ( \eta ) } { 2 ! } x ^ { 2 } \\ \\ & = f ^ { \prime } ( 0 ) x + \frac { f ^ { \prime \prime } ( \eta ) } { 2 ! } x ^ { 2 } \end{aligned} }

$$\textcolor{springgreen}{ f ( a ) = f ^ { \prime } ( 0 ) a + \frac { f ^ { \prime \prime } \left( \eta _{ 1 } \right) } { 2 ! } a ^ { 2 } } \tag{1}$$

$$\textcolor{springgreen}{ f ( – a ) = f ^ { \prime } ( 0 ) ( – a ) + \frac { f ^ { \prime \prime } \left( \eta _{ 2 } \right) } { 2 ! } a ^ { 2 } } \tag{2}$$

$$\textcolor{pink}{ f ( a ) + f ( – a ) = \frac { a ^ { 2 } } { 2 } \left[ f ^ { \prime \prime } \left( \eta _{ 1 } \right) + f ^ { \prime \prime } \left( \eta _{ 2 } \right) \right] } \tag{3}$$

\begin{aligned} m \leq f ^ { \prime \prime } \left( \eta _{ 1 } \right) \leq M \\ \\ m \leq f ^ { \prime \prime } \left( \eta _{ 2 } \right) \leq M \end{aligned}

$$m \leq \frac { f ^ { \prime \prime } \left( \eta _{ 1 } \right) + f ^ { \prime \prime } \left( \eta _{ 2 } \right) } { 2 } \leq M$$

$$\textcolor{pink}{ \frac { f ^ { \prime \prime } \left( \eta _{ 1 } \right) + f ^ { \prime \prime } \left( \eta _{ 2 } \right) } { 2 } = f ^ { \prime \prime } ( \xi )} \tag{4}$$

$$f ( a ) + f ( – a ) = a ^ { 2 } f ^ { \prime \prime } ( \xi )$$

$$f ^ { \prime \prime } ( \xi ) = \frac { f ( a ) + f ( – a ) } { a ^ { 2 } }$$

### 第 (2) 问

$$f ^ { \prime } \left( x _{ 0 } \right) = 0$$

\begin{aligned} f ( x ) \\ \\ & = f \left( x _{ 0 } \right) + f ^ { \prime } \left( x _{ 0 } \right) \left( x – x _{ 0 } \right) + \frac { f ^ { \prime \prime } ( \gamma ) } { 2 ! } \left( x – x _{ 0 } \right) ^ { 2 } \\ \\ & = f \left( x _{ 0 } \right) + \frac { f ^ { \prime \prime } ( \gamma ) } { 2 ! } \left( x – x _{ 0 } \right) ^ { 2 } \end{aligned}

$$\textcolor{pink}{ f ( – a ) = f \left( x _{ 0 } \right) + \frac { f ^ { \prime \prime } \left( \gamma _{ 1 } \right) } { 2 ! } \left( – a – x _{ 0 } \right) ^ { 2 } } \tag{5}$$

$$\textcolor{pink}{ f ( a ) = f \left( x _{ 0 } \right) + \frac { f ^ { \prime \prime } \left( \gamma _{ 2 } \right) } { 2 ! } \left( a – x _{ 0 } \right) ^ { 2 } } \tag{6}$$

\begin{aligned} | f ( a ) – f ( – a ) | \\ \\ & = \left| \frac { 1 } { 2 } \left( a – x _{ 0 } \right) ^ { 2 } f ^ { \prime \prime } \left( \gamma _{ 2 } \right) – \frac { 1 } { 2 } \left( a + x _{ 0 } \right) ^ { 2 } f ^ { \prime \prime } \left( \gamma _{ 1 } \right) \right| \\ \\ & \leq \frac { 1 } { 2 } \left| \left( a – x _{ 0 } \right) ^ { 2 } f ^ { \prime \prime } \left( \gamma _{ 2 } \right) \right| + \frac { 1 } { 2 } \left| \left( a + x _{ 0 } \right) ^ { 2 } f ^ { \prime \prime } \left( \gamma _{ 1 } \right) \right| \end{aligned}

$$M = \max \left\{ \left| f ^ { \prime \prime } \left( \gamma _{ 1 } \right) \right|, \left| f ^ { \prime \prime } \left( \gamma _{ 2 } \right) \right| \right\}$$

\begin{aligned} | f ( a ) – f ( – a ) | & \leq \frac { 1 } { 2 } M \left( a + x _{ 0 } \right) ^ { 2 } + \frac { 1 } { 2 } M \left( a – x _{ 0 } \right) ^ { 2 } \\ \\ & = M \left( a ^ { 2 } + x _{ 0 } ^ { 2 } \right) \end{aligned}

$$x _{0} \leq a$$

\begin{aligned} | f ( a ) – f ( – a ) | & \leq M \left( a ^ { 2 } + x _{ 0 } ^ { 2 } \right) \\ \\ & \leq M (a ^{2} + a ^{2}) \\ \\ & \leq 2 M a ^ { 2 } \end{aligned}

$$M \geq \frac { 1 } { 2 a ^ { 2 } } | f ( a ) – f ( – a ) |$$

$$\left| f ^ { \prime \prime } ( \eta ) \right| \geq \frac { 1 } { 2 a ^ { 2 } } | f ( a ) – f ( – a ) |$$