Question

find the maximum and minimum points of the function x³-x²

Answer 1

$$y=x^3-x^2$$

Domain: $(-\infin, \infin)$

Find the first derivative

$$y'=(x^3-x^2)'=3x^2-2x$$

Find the critical point(s)

$$y'=0=>3x^2-2x=0$$

$$x(3x-2)=0$$

$$x_1=0, x_2=\dfrac{2}{3}$$

Critical numbers: $0, \dfrac{2}{3}.$

First Derivative Test

If $x<0,$ then $y'>0, y$ increases.

If $0<x<\dfrac{2}{3},$ then $y'<0, y$ decreases.

If $x>\dfrac{2}{3},$ then $y'>0, y$ increases.

$$y(0)=(0)^3-(0)^2=0$$

$$y(0)=(\dfrac{2}{3})^3-(\dfrac{2}{3})^2=-\dfrac{4}{27}$$

The function $y=x^3-x^2$ has a local maximum with value of at $x=0,$

$Point(0, 0).$

The function $y=x^3-x^2$ has a local minimum with value of $\dfrac{2}{3}$ at $x=-\dfrac{4}{27},$

$Point(-\dfrac{2}{3}, -\dfrac{4}{27}).$

Since the function $y=x^3-x^2$ is defined on $\R,$ the function $y$ has neither absolute maximum nor absolute minimum.