计算一阶导数：

识别临界点：$f^{\prime }(x)$对所有实数$x$都有定义，因此令$f^{\prime }(x)=0$得到临界点$x=1$和$x=3$，两者都在$f$定义域内部。

符号分析：将定义域分为$(-\infty ,1)$, $(1,3)$, $(3,\infty )$，在每个区间测试一个点：

• At $x=0$: $f^{\prime }(0)=9>0$
  - At $x=2$: $f^{\prime }(2)=-3<0$
  - At $x=4$: $f^{\prime }(4)=9>0$

根据符号变化分类：在$x=1$处，$f^{\prime }$由正变负 → $(1,6)$处为相对极大值。在$x=3$处，$f^{\prime }$由负变正 → $(3,2)$处为相对极小值。

使用乘积法则计算$f^{\prime }(x)$：

识别临界点：$f^{\prime }(x)$在$x=0$处不存在，且$f(0)=0$（因此$x=0$在$f$的定义域中），所以它是一个有效的临界点。令分子等于零得到$x=8/5$，此处$f^{\prime }=0$且$f$有定义，因此$x=8/5$也是一个临界点。

对区间$(-\infty ,0)$, $(0,8/5)$, $(8/5,\infty )$进行符号分析：

• At $x=-1$: $f^{\prime }(-1)=\frac{-13}{-3}>0$
  - At $x=1$: $f^{\prime }(1)=\frac{-3}{3}<0$
  - At $x=2$: $f^{\prime }(2)=\frac{2}{3\cdot 2^{1/3}}>0$

分类：在$x=0$处，$f^{\prime }$由正变负 → $(0,0)$处为相对极大值。在$x=8/5$处，$f^{\prime }$由负变正 → $(1.6,\approx -3.28)$处为相对极小值。

Find interior critical points: $f^{\prime }(x)=6x^2-6x=6x(x-1)$, so interior critical points at $x=0$ and $x=1$.

Sign analysis for interior points: Intervals $(-1,0)$, $(0,1)$, $(1,2)$. Testing gives $f^{\prime }(-0.5)>0$, $f^{\prime }(0.5)<0$, $f^{\prime }(1.5)>0$. So $x=0$ is a relative maximum, $x=1$ is a relative minimum.

Classify endpoints: Left endpoint $x=-1$ has $f^{\prime }(x)$ positive just right of $-1$, so $f$ increases away from $x=-1$ → relative minimum at $(-1,-5)$. Right endpoint $x=2$ has $f^{\prime }(x)$ positive just left of $2$, so $f$ increases towards $x=2$ → relative maximum at $(2,4)$.

Final result: Relative minima at $(-1,-5)$ and $(1,-1)$; relative maxima at $(0,0)$ and $(2,4)$.

(a) Use the quotient rule to compute the first derivative:

The domain of $f$ excludes $x=2$, so even though $f^{\prime }$ is undefined at $x=2$, it is not a critical point. $f^{\prime }(x)=0$ at $x=0$ and $x=4$, both in the domain of $f$, so critical points are $x=0$ and $x=4$.

(b) The denominator $(x-2{)}^2$ is always positive for $x\ne 2$, so the sign of $f^{\prime }$ matches the sign of $x(x-4)$. Testing intervals: $f^{\prime }>0$ for $x<0$, $f^{\prime }<0$ for $0<x<2$ and $2<x<4$, and $f^{\prime }>0$ for $x>4$. At $x=0$, $f^{\prime }$ changes from positive to negative → relative maximum. At $x=4$, $f^{\prime }$ changes from negative to positive → relative minimum.

(c) On $[0,4]$, endpoints are $x=0$ and $x=4$, which we already classified, and $x=2$ is not in the domain. The relative extrema are: relative maximum at $(0,0)$ and relative minimum at $(4,8)$.

Compute the first derivative:

Find critical points: Set $P^{\prime }(x)=0$, which simplifies to $x^2-80x-1000=0$. Solving with the quadratic formula gives one positive critical point at $x\approx 91$, which is inside the interval $[0,100]$.

Sign analysis: Test $x=10$: $P^{\prime }(10)=5.1>0$. Test $x=95$: $P^{\prime }(95)\approx -1.28<0$.

Classification: $P^{\prime }$ changes from positive to negative at $x\approx 91$, so this is a relative maximum. $P(91)\approx 482$ dollars.

Interpretation: Producing approximately 91 units per day gives the business a relative maximum daily profit of roughly $482.