Question

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

Answer 1

Below

Explanation:

`y=1/3x^3-2x^2+3x+2`
`(dy)/(dx)=x^2-4x+3`

For turning points, `(dy)/(dx)=0`

`x^2-4x+3=0`
`(x-3)(x-1)=0`
`x=3` or `x=1`

To see whether your point is a maximum or minimum, we use the second derivative
`(d^2y)/(dx^2)=2x-4`

When `x=3`,
`(d^2y)/(dx^2)=2(3)-4`
`(d^2y)/(dx^2)=2` which is greater than 0 so it is a minimum point
`(3,2)` is a minimum point

When `x=1`,
`(d^2y)/(dx^2)=2(1)-4`
`(d^2y)/(dx^2)=-2` which is less than 0 so it is a maximum point
`(1,3 1/3)` is a maximum point

Below is the graph
graph{y=1/3x^3-2x^2+3x+2 [-18.02, 18.02, -9.01, 9.01]}