曲线 $C$ 的方程为 $y=\dfrac{x^2+x+1}{x+1}$。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上所有驻点的坐标。

(c) 画出 $C$ 的草图。

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

(e) 求满足 $\dfrac{|x|^2+|x|+1}{|x|+1}<3$ 的 $x$ 的值的集合，以精确形式表示。

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。

多项式除法拆出斜线与余项，并核对分母零点没有被约去。

Since $x^2+x+1=x(x+1)+1$,

The numerator equals $1$ at $x=-1$, so the vertical asymptote is $\boxed{x=-1}$. Since $1/(x+1)\to0$ as $|x|\to\infty$, the oblique asymptote is $\boxed{y=x}$.

对拆分式求导，解导数为零并代回函数。

Differentiating,

Setting this equal to zero gives $(x+1)^2=1$, so $x=0$ or $x=-2$. Substitution gives the stationary points

The derivative changes from positive to negative at $-2$ and from negative to positive at $0$, so these are a local maximum and local minimum, respectively.

标两渐近线与驻点，两支分别位于斜渐近线下方和上方。

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

• For $x<-1$, the branch lies below $y=x$, rises to the smooth maximum $(-2,-3)$, then tends to $-\infty$ as $x\to-1^-$.
• For $x>-1$, it lies above $y=x$, tends to $+\infty$ as $x\to-1^+$, falls to the smooth minimum $(0,1)$ and then increases towards the oblique asymptote.

There are no $x$-intercepts because $x^2+x+1>0$ for all real $x$.

把输入x换成绝对值：保留原图x非负的部分，并关于纵轴反射。不是把横轴下方的函数值取绝对值。

Write $\mathrm f(x)=(x^2+x+1)/(x+1)$. The new graph is $y=\mathrm f(|x|)$. Keep the part of the original graph with $x\ge0$ and reflect it in the $y$-axis.

The result is symmetric about $x=0$, has a smooth minimum at $(0,1)$ and no vertical asymptotes. Its oblique asymptotes are $y=x$ on the right and $y=-x$ on the left. The curve lies above these asymptotes.

设u为|x|，则u非负且分母恒正。先解关于u的二次不等式，再换回x的区间。

Put $u=|x|\ge0$. Since $u+1>0$,

The roots are $1\pm\sqrt3$. Intersecting the interval between them with $u\ge0$ gives $0\le u<1+\sqrt3$. Therefore