Find the extremum (1) 5x-3x ^ 2 (2) 2x ^ 2 + x-3 (3) 1 / (x ^ 2 + X + 1) (4) x ^ 2 + 4 / x ^ 2 of the following formula. Thank you very much

Find the extremum (1) 5x-3x ^ 2 (2) 2x ^ 2 + x-3 (3) 1 / (x ^ 2 + X + 1) (4) x ^ 2 + 4 / x ^ 2 of the following formula. Thank you very much

(1)
-3x^2 + 5x = -3 (x-5/6)^2 + 25/12
When x = 5 / 6, there is a maximum of 25 / 12
(2)
2x^2 + x - 3 = 2(x+1/4)^2 - 25/8
When x = - 1 / 4, there is a minimum of - 25 / 8
(3)
Because x ^ 2 + X + 1 = (x + 1 / 2) ^ 2 + 3 / 4
So 1 / (x ^ 2 + X + 1) has a maximum of 4 / 3 when x = - 1 / 2
(4)
x^2 + 4/x^2 >= 2*x*(2/x) = 4
When x ^ 2 = 4 / x ^ 2, that is, x = positive and negative root 2, there is a minimum value of 4