Find the critical points, and determine the maximum and minimum.g(y)=(y-1)/(y^2-y+1)
`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`
Set each factor to zero and solve for y.
`y =0` and ` 2-y=0`
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) `