for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where…

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.\n1) $y =-\frac{x^{3}}{3}+x^{2}$

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.\n1) $y =-\frac{x^{3}}{3}+x^{2}$

Answer

Explanation:

Step1: 求x - 截距

令$y = 0$，即$-\frac{x^{3}}{3}+x^{2}=0$，提取公因式$x^{2}$得$x^{2}(1 - \frac{x}{3})=0$，解得$x = 0$或$x = 3$。

Step2: 求y - 截距

令$x = 0$，则$y=0$。

Step3: 求一阶导数

对$y =-\frac{x^{3}}{3}+x^{2}$求导，$y^\prime=-x^{2}+2x$。

Step4: 求临界值

令$y^\prime = 0$，即$-x^{2}+2x = 0$，提取公因式$x$得$x(2 - x)=0$，解得$x = 0$或$x = 2$。

Step5: 确定单调区间

当$y^\prime>0$时，函数递增，$-x^{2}+2x>0$，即$x(2 - x)>0$，解得$0<x<2$，递增区间为$(0,2)$；当$y^\prime<0$时，函数递减，解得$x<0$或$x>2$，递减区间为$(-\infty,0)\cup(2,+\infty)$。

Step6: 求二阶导数

对$y^\prime=-x^{2}+2x$求导，$y^{\prime\prime}=-2x + 2$。

Step7: 求拐点

令$y^{\prime\prime}=0$，即$-2x + 2=0$，解得$x = 1$。

Step8: 确定凹凸区间

当$y^{\prime\prime}>0$时，函数上凹，$-2x + 2>0$，解得$x<1$，上凹区间为$(-\infty,1)$；当$y^{\prime\prime}<0$时，函数下凹，解得$x>1$，下凹区间为$(1,+\infty)$。

Step9: 求极值

因为$y^\prime$在$x = 0$左侧小于0，右侧大于0，所以$x = 0$处为极小值点，$y(0)=0$；$y^\prime$在$x = 2$左侧大于0，右侧小于0，所以$x = 2$处为极大值点，$y(2)=-\frac{8}{3}+4=\frac{4}{3}$。

Answer:

x - 截距：$x = 0,x = 3$；y - 截距：$y = 0$； 临界值：$x = 0,x = 2$； 递增区间：$(0,2)$；递减区间：$(-\infty,0)\cup(2,+\infty)$； 拐点：$x = 1$； 上凹区间：$(-\infty,1)$；下凹区间：$(1,+\infty)$； 极小值：$y(0)=0$；极大值：$y(2)=\frac{4}{3}$ 。