已知函数f(x)=ax²+bx+1在【-2,a】上是偶函数,则f(x)=?
人气:444 ℃ 时间:2019-11-18 06:06:49
解答
因为偶函数的定义域关于原点对称,所以a=2.
因为f(x)的对称轴为x=-b/(2a),偶函数的对称轴为x=0,所以b=0
f(x)=2x²+1.
推荐
- 已知函数f(x)=ax²+bx+c,x属于[-2a-3,1]是偶函数,求a+b的值
- 已知二次函数f(x)=ax²+bx+1为偶函数,且f(-1)=-1
- 已知f(x)=ax²+bx+3a+b是偶函数,定义域为〔a-1,2a〕,则f(1/2)是多少
- 已知二次函数f(x)=ax²+bx+1是偶函数,且f(1)=0.(1)求a,b (2)设g(x)=f(x+2).若g(x)在区间[-2,m]上的最小值为-3,求实数m的值
- 已知f(x)=ax²+bx(a≠0,b≠R),且y=f(x+1)为偶函数,方程f(x)=x有两个相等的实数根,求函数 f(x)的解析式
- 过去式 they were sang 还是they was sang 和 ...he were slept 还是 he was slept?.谢
- “This is ______ most useful reference book,” a teacher from ______ European country told us in cl
- he makes toys in the factory哪里错了
猜你喜欢