Question

Sketch and determine the turning points of the curve Y=1/3x^3-2x^2+3x+2?

Answer 1

The local maximum is at `(1,10/3)` and the local minimum is at `(3,2)`

Explanation:

The function is

`y=1/3x^3-2x^2+3x+2`

The first derivative is

`dy/dx=x^2-4x+3`

The critical points are when `dy/dx=`

That is,

`x^2-4x+3=0`

`(x-3)(x-1)=0`

`x=3` and `x=1`

We can build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `1` `color(white)(aaaaaaaa)` `3` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaa)` `-` `color(white)(aa)` ###color(white)(aaaa)#`-` `color(white)(aa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `dy/dx` `color(white)(aaaaaa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaa)` `↗` `color(white)(aa)` ###color(white)(aaaa)#`↘` `color(white)(aa)` ###color(white)(aaaaa)#`↗`

We calculate the second derivative

`(d^2y)/(dx^2)=2x-4`

When `(d^2y)/(dx^2)=0`, there is a point of inflection

`2x-4=0`, `=>`, `x=2`

We can build a variation chart with the second derivative

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,2)` `color(white)(aaaa)` `(2,+oo)`

`color(white)(aaaa)` `sign (d^2y)/dx^2` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaaaaaa)` `nn` `color(white)(aaaaaaaaa)` `uu`

graph{1/3x^3-2x^2+3x+2 [-14.24, 14.24, -7.12, 7.12]}