曲线 $C$ 的方程为 $y=\mathrm{f}(x)$，其中

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上所有驻点的坐标，答案精确到小数点后一位。

(c) 画出 $C$ 的草图。

(d) 画出方程为 $y=\dfrac{1}{\mathrm{f}(x)}$ 的曲线的草图，并写出满足 $\dfrac{1}{\mathrm{f}(x)}<\mathrm{f}(x)$ 的值的集合。

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。

分母有两个实零点且分子在两处都非零，因此有两条竖直渐近线。分子分母最高次项系数相同，水平渐近线为1。

The denominator is $2x(x-1)$. The numerator equals $1$ at $x=0$ and $2$ at $x=1$, so no factors cancel. Therefore the vertical asymptotes are

The ratio of the leading coefficients is $2/2=1$, so the horizontal asymptote is $\boxed{y=1}$.

商法则求导后令分子为零，求精确横坐标并代回，最后按题目要求保留一位小数。

By the quotient rule,

Thus the stationary points satisfy

The corresponding values are

Therefore, to $1$ decimal place, the stationary points are

先用函数减1确定与水平渐近线的交点及相对位置，再结合两条竖直渐近线画三支。

Since

the curve crosses the horizontal asymptote at $(-1,1)$. The numerator of $\mathrm f$ is always positive because its discriminant is $1-8=-7<0$, so there are no $x$-intercepts.

Sketch three branches, labelling $x=0$, $x=1$, $y=1$:

• On $x<0$, the branch approaches $y=1$ from below as $x\to-\infty$, has a minimum near $(-2.4,0.9)$, crosses $(-1,1)$ and tends to $+\infty$ as $x\to0^-$.
• On $0<x<1$, the branch is negative, tends to $-\infty$ at both ends and has a maximum near $(0.4,-1.9)$.
• On $x>1$, the branch decreases from $+\infty$ to $y=1$ from above.