You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation g'(y) = 0.

You need to evaluate the first derivative, using the quotient rule:

`g'(y)= ((y-1)'(y^2-y+1) - (y-1)(y^2-y+1)')/((y^2-y+1)^2)`

`g'(y)= (y^2-y+1 - (y-1)(2y-1))/((y^2-y+1)^2)`

`g'(y)= (y^2-y+1 - 2y^2 + 3y - 1))/((y^2-y+1)^2)`

`g'(y)= (-y^2+2y)/((y^2-y+1)^2)`

You need to solve for g'(y) = 0, such that:

`-y^2+2y = 0 => y^2 - 2y = 0`

Factoring out y yields:

`y(y - 2) = 0 => y = 0`

`y - 2 = 0 => y = 2`

**Hence, evaluating the critical numbers of the function for g'(y) = 0, yields y = 0 and y = 2.**

Critical values would be where there are

- Endpoints
- f’(x)=0
- f’(x) DNE

`g'(y)=((1)(y^2-y+1)-(y-1)(2y-1))/(y^2-y+1)`

`g'(y)=-((y-2)y)/(y^2-y+1)^2 `

First and third rules don't apply. Set g'(y)=0

`y(-y+2)=0 `

`y=0 ` and `y=2 `