2021年考研数二第18题解析：凹凸性、渐近线

一、题目

难度评级：

二、解析

首先，由 $f\left(x\right) = \dfrac{x\left|x\right|}{1+x}$, 可得：

$$
f\left(x\right) = \begin{cases} \frac{x^{2}}{1+x}, & x>0 \\ -\frac{x^{2}}{1+x}, & x<0 \end{cases}
$$

由于判断函数的凹凸性需要用到二阶导函数，于是——

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $x>0$ 时：

$$
\begin{aligned}
f^{\prime}\left(x\right) & = \frac{2x+x^{2}}{\left(1+x\right)^{2}} \\ \\
f^{\prime\prime}\left(x\right) & = \frac{2}{\left(1+x\right)^{3}}
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $x<0$ 时：

$$
\begin{aligned}
f^{\prime}\left(x\right) & = \frac{-2x-x^{2}}{\left(1+x\right)^{2}} \\ \\
f^{\prime\prime}\left(x\right) & = \frac{-2}{\left(1+x\right)^{3}}
\end{aligned}
$$

所以：

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
x & \left(-\infty,-1\right) & -1 & \left(-1,0\right) & 0 & \left(0,+\infty\right) \\
\hline
f^{\prime\prime}\left(x\right) & + & & – & & + \\
\hline
f\left(x\right) & \text{凹} & \text{拐点} & \text{凸} & \text{拐点} & \text{凹} \\
\hline
\end{array}
$$

于是可知：

$\textcolor{lightgreen}{\blacktriangleright}$ 凹区间为：

$$
\left(-\infty,-1\right), \ \left(0,+\infty\right)
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 凸区间为：

$$
\left(-1,0\right)
$$

接下来判断渐近线的类型——

$\textcolor{lightgreen}{\blacktriangleright}$ 垂直渐近线

$\lim_{x \to -1}\frac{x\left|x\right|}{1+x}=\infty$，$x=-1$ 是垂直渐近线.

$\textcolor{lightgreen}{\blacktriangleright}$ 水平渐近线

由于：

$$
\begin{aligned}
\lim_{x \to + \infty} \frac{x\left|x\right|}{1+x} & = \lim_{x \to + \infty} \frac{x^{2}}{x} = \lim_{x \to + \infty} x \to + \infty \\ \\
\lim_{x \to – \infty} \frac{x\left|x\right|}{1+x} & = \lim_{x \to – \infty} \frac{-x^{2}}{x} = \lim_{x \to – \infty} x \to – \infty
\end{aligned}
$$

由于极限 $\lim_{x \to + \infty} \frac{x\left|x\right|}{1+x}$ 和极限 $\lim_{x \to – \infty} \frac{x\left|x\right|}{1+x}$ 都不存在，所以，函数 $f\left(x\right) = \dfrac{x\left|x\right|}{1+x}$ 没有水平渐近线.

$\textcolor{lightgreen}{\blacktriangleright}$ 倾斜渐近线

由于：

$$
\begin{aligned}
k & = \lim_{x \to +\infty} \frac{f(x)}{x} \\ \\
& = \lim_{x \to +\infty} \frac{x\left|x\right|}{\left(1+x\right)x} \\ \\
& = \lim_{x \to +\infty} \frac{x^{2}}{x^{2}} \\ \\
& = 1 \\ \\ \\
b & = \lim_{x \to +\infty} \left[ f(x) \textcolor{pink}{- k } x \right] \\ \\
& = \lim_{x \to +\infty} \left[\frac{x\left|x\right|}{1+x} \textcolor{pink}{-} x \right] \\ \\
& = \lim_{x \to +\infty} \left[\frac{x^{2}}{1+x} – x \right] \\ \\
& = \lim_{x \to +\infty} \left[\frac{x^{2}}{1+x} – \frac{x \left( 1+x \right)}{1+x} \right] \\ \\
& = \lim_{x \to +\infty} \frac{-x}{1+x} \\ \\
& = \lim_{x \to +\infty} \frac{-x}{x} \\ \\
& = -1
\end{aligned}
$$

因此，存在斜渐近线：

$$
\textcolor{lightgreen}{
y = kx+b = x-1
}
$$

又由于：

$$
\begin{aligned}
k & = \lim_{x \to -\infty} \frac{f(x)}{x} \\ \\
& = \lim_{x \to -\infty} \frac{x\left|x\right|}{\left(1+x\right)x} \\ \\
& = \lim_{x \to +\infty} \frac{-x^{2}}{x^{2}} \\ \\
& = -1 \\ \\ \\
b & = \lim_{x \to -\infty} \left[ f(x) \textcolor{orangered}{- k} x \right] \\ \\
& = \lim_{x \to -\infty} \left[\frac{x\left|x\right|}{1+x} \textcolor{orangered}{+} x \right] \\ \\
& = \lim_{x \to -\infty} \left[\frac{-x^{2}}{1+x} + x \right] \\ \\
& = \lim_{x \to -\infty} \left[\frac{-x^{2}}{1+x} + \frac{x \left( 1+x \right)}{1+x} \right] \\ \\
& = \lim_{x \to -\infty} \frac{x}{1+x} \\ \\
& = \lim_{x \to -\infty} \frac{x}{x} \\ \\
& = 1
\end{aligned}
$$

因此，存在斜渐近线：

$$
\textcolor{lightgreen}{
y = kx+b = -x+1
}
$$

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