一条曲线的方程为 $y=\mathrm{f}(x)$，其中 $\mathrm{f}(x)=(4x-3)^{\frac{5}{3}}-\dfrac{20}{3}x$。

(a) 求曲线各驻点的 $x$ 坐标，并确定它们的类型。

(b) 写出函数 $\mathrm{f}$ 递增时所对应的取值集合。

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。

把分数幂理解为实立方根的幂。求导后用立方根的平方解驻点方程，并检查导数变号。

The derivative is

It is zero when $\sqrt[3]{4x-3}=\pm1$, giving $x=1/2$ and $x=1$. Since $(4x-3)^{2/3}=|4x-3|^{2/3}$, the derivative is positive outside $(1/2,1)$ except at the endpoints, and negative inside. Hence $x=1/2$ gives a local maximum and $x=1$ gives a local minimum.

求出同样的两个驻点后，也可代入二阶导数，分别判断曲线在这两点向下或向上弯曲。

The equation $\mathrm f'(x)=0$ gives $(4x-3)^{2/3}=1$, hence $4x-3=\pm1$ and $x=1/2$ or $1$. For $x\ne3/4$,

At $x=1/2$, this is $-160/9<0$, so the stationary point is a local maximum. At $x=1$, it is $160/9>0$, so the stationary point is a local minimum.

沿用 (a) 的导数符号，分别写出两个递增区间。

The derivative is positive for $x<1/2$ or $x>1$, so these are the two open intervals of increase. Either endpoint may also be included in its corresponding interval, as the zero derivative there does not prevent strict increase on that interval.