# Find the critical points, and determine the maximum and minimum.g(y)=(y-1)/(y^2-y+1)

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

(a) To determine the critical points, take the derivative of g(y) with respect to y. Use the quotient formula of derivatives which is `(u/v)' = (vu'-uv')/v^2`

So we have,

`u = y-1 ` `v = y^2-y+1`

`u' = 1 ` `v' = 2y - 1`

Substitute these to the formula.

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

Set g'(y) equal to zero.

`0 = (-y^2+2y)/(y^2-y+1)^2`

`0 = -y^2+2y`

`0 =y(2-y)`

Set each factor to zero and solve for y.

`y =0` and ` 2-y=0`

`y=2`

Then, substitute values of y to g(y).

`y=0` , `g(y) = (0-1)/(0^2-0+1) = -1`

`y=2` , `g(y) = (2-1)/(2^2-2+1) = 1/(4-2+1)=1/3`

*Hence, the critical points are `(0,-1)` and `(2,1/3)` .*

(b) Since there are only two critical points, then the maximum point is (2,1/3) and the minimum point is (0,-1).

To verify, set the critical values x=0 and x=1 as boundaries of our intervals. Then, assign any values of x for each interval. Substitute it to g'(y) to be able to determine the sign of g'(y).

For Interval I (x<0),

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

For Interval II (0<x<2),

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

For Interval III (x>2),

let `x=3` , `g'(y) = (-3^2+2(3))/(3^2-3+1)^2 = -3/7^2 = -3/49`

Since the change of sign from interval I to II is negative(-) to positive (+), then the critical point between these two intervals, which is *`(0,-1)` is the minimum point*.

And *`(2,1/3)` is the maximum point* since the change of signs of the interval beside it is from (+) to (-).

**Answer: Critical points: `(0,-1)` and `(2,1/3)` **

** Minimum point: `(0,-1)` **

** Maximum point: `(2,1/3) ` **