一条曲线的方程为 $y=3x+1-4(3x+1)^{\frac12}$，其中 $x>-\frac13$。

(a) 求 $\frac{\mathrm{d}y}{\mathrm{d}x}$ 和 $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$。

(b) 求曲线驻点的坐标，并判断其性质。

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。

两次求导都要乘以内层 $3x+1$ 的导数。把负次幂保留下来便于第二次求导。

The chain rule gives

令一阶导数为零求横坐标，再用二阶导数判定性质。

The stationary condition gives $\sqrt{3x+1}=2$, so $x=1$. The height is $4-4(2)=-4$. At $x=1$, the second derivative is $9/8>0$, so $(1,-4)$ is a local minimum.

也可用一阶导数在驻点两侧的符号判断。根式在定义域内为正，方便直接比较它与2的大小。

Setting $3-6/\sqrt{3x+1}=0$ gives $\sqrt{3x+1}=2$, hence $x=1$ and $y=-4$. For $-1/3<x<1$, the positive square root is less than $2$, so $\mathrm{d}y/\mathrm{d}x<0$. For $x>1$, it is greater than $2$, so $\mathrm{d}y/\mathrm{d}x>0$. Thus the curve changes from decreasing to increasing at $(1,-4)$, which is a local minimum.