问题
填空题
函数f(x)=x2+2x-1(x∈R)的值域是______.
答案
对函数式进行配方得到:y=x2+2x-1=(x+1)2-2,
∵函数的定义域是R,于是可得函数的最小值为-2,从而函数的值域为:[-2,+∞).
故答案为:[-2,+∞).
函数f(x)=x2+2x-1(x∈R)的值域是______.
对函数式进行配方得到:y=x2+2x-1=(x+1)2-2,
∵函数的定义域是R,于是可得函数的最小值为-2,从而函数的值域为:[-2,+∞).
故答案为:[-2,+∞).
Passage 3
Which of the following is not mentioned as the products of 3M Co. in this news report a. Optical film. b. Post-it notes. c. Scotch tape. d. Desktops.