已知函数 $f(x)=xe^x+ax+b$，曲线 $y=f(x)$ 在 $(0,f(0))$ 处的切线方程为 $y=-2x+1$。

3. 当 $x>0$ 时，$f(x+\boldsymbol{k})+f(\boldsymbol{k}-x)>2f(\boldsymbol{k})$，求 $\boldsymbol{k}$ 的最小值。

第三问详解

解题大纲

1. 构造 $\large h(x)$，计算$\large h(0)$
2. 计算 $\large h'(x)$和$\large h'(0)$
3. 证明要满足题目条件，需满足 $\large h''(0)\geq 0$
4. 计算 $\large h''(x),h''(0)$
5. 求得当 $\large k\geq -2$时，$\large h''(0)\geq 0$
6. 证明当 $\large k=-2$时，$\large h(x) > 0$恒成立，得到答案$\large k$的最小值为$\large -2$

令 $\large h(x) = f(x+k)+f(k-x)-2f(k)$

当 $\large h(x)\gt 0$恒成立时，满足题目要求

$\large h(0)=f(k)+f(k)-2f(k) = 0$

$$\begin{aligned} f'(x) &= (x)'e^{x}+x(e^{x})' - 3\\ &=(x+1)e^{x} \end{aligned}$$

$$\begin{aligned} h'(x)&=f'(x+k)\colorbox{yellow}{\((x+k)'\)}+f'(k-x) \colorbox{yellow}{\((k-x)'\)}\\ &=f'(x+k)-f'(k-x) \\ &=\big[(x+k+1)e^{x+k}-3\big]-\big[(k-x+1)e^{k-x}-3\big] \\ &=(x+k+1)e^{x+k}-(k-x+1)e^{k-x} \end{aligned}$$

$\large h'(0) = (k+1)e^k - (k+1)e^k = 0$

引理：满足题目条件的必要条件为 $\large h''(0) \geq 0$

证明：如果 $\large h''(0) \lt 0$，又 $\large \because h'(0) = 0$

$\large \therefore$，存在 $\large a$使得 $\large x \in (0,a)$时，$\large h'(x) \lt 0$

又 $\large \because h'(0) = 0,h(0) = 0$，存在 $\large b$使得 $\large x \in (0,b)$时，$\large h(x) \lt 0$，不满足题目条件

$\large \therefore$满足题目条件的必要条件为 $\large h''(0) \geq 0$

$$\begin{aligned} f''(x)&=[(x+1)e^x-3]' \\ &=(x+1)'e^x + (x+1)(e^x)' \\ &=(x+2)e^x \end{aligned}$$

$$\begin{aligned} h'(x)&=f'(x+k)(x+k)' + f'(k-x)(k-x)' \\ &=f'(x+k)-f'(k-x) \end{aligned}$$

$$\begin{aligned} h''(x) &= f''(x+k)(x+k)'-f''(k-x)(k-x)' \\ &=f''(x+k)+f''(k-x) \\ &=(x+k+2)e^{x+k}+(k-x+2)e^{k-x} \end{aligned}$$

$h''(0) = 2(k+2)e^k$

$$\begin{aligned} &\qquad \quad 2(k+2)e^k \geq 0 \\ &\implies {\Large \colorbox{yellow}{\(k\geq -2\)} } \end{aligned}$$

当$\large k=-2$时，$\large h''(0) = 0$

$$\begin{aligned} h''(x)&= (x)e^{x-2}+(-x)e^{-2-x} \\ &=e^{-2}x[e^x-e^{-x}] \\ \end{aligned}$$

当 $\large x\gt 0时$，$\large e^x > 1 > e^{-x}$

$\large \therefore h''(x)恒大于0$

$\large \therefore h'(x)恒大于0$

$\large \therefore h(x)恒大于0$，满足条件

$\large \colorbox{yellow}{\(k的最小值为-2\)}$