曲线 $C$ 的方程为 $y=\dfrac{x+1}{x^2+x+1}$。

(a) 证明 $C$ 没有竖直渐近线，并写出水平渐近线的方程。

(b) 求 $C$ 上所有驻点的坐标。

(c) 画出 $C$ 的草图，并写出它与坐标轴交点的坐标。

(d) 画出方程为 $y=\bigg|\dfrac{x+1}{x^2+x+1}\bigg|$ 的曲线的草图。

(e) 求满足 $\bigg|\dfrac{x+1}{x^2+x+1}\bigg|<\frac{1}{3}$ 的 $x$ 值的集合，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。

分母配方后恒正，所以没有实数零点，也就没有竖直渐近线。分母次数较高给出水平渐近线。

The denominator satisfies

for every real $x$. It never vanishes, so $C$ has no vertical asymptotes.

Also

Thus the horizontal asymptote is $\boxed{y=0}$.

用商法则求导，分母恒正，所以驻点由导数分子的零点确定。

By the quotient rule,

Setting the derivative to zero gives $x=-2$ or $x=0$. Substitution gives the stationary points

The derivative is negative for $x<-2$, positive for $-2<x<0$, and negative for $x>0$. Hence these are a local minimum and a local maximum, respectively.

标出两轴交点、两个驻点以及从上下两侧趋近的水平渐近线。

The intercepts are $\boxed{(-1,0)}$ and $\boxed{(0,1)}$.

Sketch a continuous curve for all real $x$, approaching $y=0$ from below as $x\to-\infty$, falling to the minimum $(-2,-1/3)$, then rising through $(-1,0)$ to the maximum $(0,1)$. It then decreases towards $y=0$ from above as $x\to+\infty$.

只反射原曲线在横轴下方的部分；原横轴交点变成尖点，不能画成光滑极小值。

Reflect the portion with $x<-1$ in the $x$-axis and leave the portion with $x\ge-1$ unchanged. The minimum $(-2,-1/3)$ becomes a smooth maximum $(-2,1/3)$.

The transformed graph has a sharp minimum at $(-1,0)$, retains the maximum $(0,1)$ and approaches $y=0$ from above at both ends.

分母恒正，将绝对值不等式化为同时满足的上下界。下界会排除一个孤立点，不能漏掉。

Since $x^2+x+1>0$, the inequality is equivalent to

The upper bound gives

so $x<1-\sqrt3$ or $x>1+\sqrt3$.

The lower bound gives

which holds for every real $x$ except $x=-2$.

Combining these conditions,