曲线 $C$ 的方程为 $y=\mathrm{f}(x)$，其中 $\mathrm{f}(x)=\dfrac{x^2}{x+1}$。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上所有驻点的坐标。

(c) 画出 $C$ 的示意图。

(d) 求方程为 $y=\dfrac{1}{\mathrm{f}(x)}$ 的曲线上所有驻点的坐标。

(e) 画出方程为 $y=\dfrac{1}{\mathrm{f}(x)}$ 的曲线的示意图，并求满足 $\dfrac{1}{\mathrm{f}(x)}>\mathrm{f}(x)$ 的取值集合，用精确形式表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先除多项式，得到斜渐近线与余项；分母零点处没有约消。

Division gives

Therefore the asymptotes are $\boxed{x=-1}$ and $\boxed{y=x-1}$.

由除法式求导，分子因式分解后得到两个驻点。

Thus $f'(x)=0$ when $x=0$ or $x=-2$, both in the domain. Substitution gives the stationary points

The derivative changes from positive to negative at $x=-2$ and from negative to positive at $x=0$, giving a maximum and minimum, respectively.

用余项符号决定曲线位于斜渐近线哪一侧。左支最高点仍在横轴下，右支最低点恰在原点。

Sketch the two branches with labelled asymptotes $x=-1$ and $y=x-1$:

• For $x<-1$, the curve lies below the oblique asymptote, rises to the maximum $(-2,-4)$ and then falls to $-\infty$ as $x\to-1^-$.
• For $x>-1$, it lies above the oblique asymptote, falls from $+\infty$ as $x\to-1^+$ to the minimum $(0,0)$, touches the $x$-axis there, and then increases to $+\infty$.

Both tails approach the oblique asymptote; $(0,0)$ is the only intersection with either coordinate axis.

先保留倒数的共同定义域，再求化简式的导数。分子零点必须回查定义域。

The reciprocal is defined for $x\ne-1,0$. On this domain,

Hence

The only zero is $x=-2$, which is in the domain, and $g(-2)=-1/4$. Thus the only stationary point is

不等式部分先移到同一边并因式分解。x平方和另一个二次式恒正，只剩一个二次因子与x+1的符号决定答案，另排除倒数无定义的0。

For the inequality, work on the common domain $x\ne-1,0$. Then

Since $x^2>0$ on the domain and $x^2+x+1=(x+1/2)^2+3/4>0$, the required expression is positive exactly when

The numerator roots are $a=(1-\sqrt5)/2$ and $b=(1+\sqrt5)/2$, with $-1<a<0<b$.

Therefore the inequality holds for