《中学生数学》2013年第4月(下)课外练习题初三年级第1题是:题求函数y=2x+2x2+3x+3的最大值和最小值.参考答案用“判别式”给出了解答.本文再给出一种不用“判别式”的解法,供同学们参阅.另解y=2x+2x2+3x+3=2(x+1)(x+1)2+(x+1)+1,当x+1=0时,y=0,即y=0是函数的一个值;当x+1≠0时,y=2x+2x2+3x+3=2(x+1)(x+1)2+(x+1)+1=2(x+1)+1x+1+1.∵|x+1|+1|x+1|≥2|x+1|·1|x+1槡|=2, “Mathematics for secondary school students” in April 2013 (below) Extracurricular exercises The first three grades of the first question is: the maximum and minimum values ​​of the function y = 2x + 2x2 + 3x + 3. Reference answer with “discriminant ”Gives the answer. This article gives a solution without “ discriminant ”for the students to see. Another solution y = 2x + 2x2 + 3x +3 = 2 (x +1) (x +1) (X + 1) +1, y = 0 when x + 1 = 0, that is, y = 0 is a value of the function; y = 2x + 2x2 + 3x + 3 = 2 x + 1 ⁢ x + 1 + 2 + x + 1 = 2 ⁢ x + 1 + 1x + 1 + x + 1 | · 1 | x + 1 槡 | = 2,