设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件 (1) 当x∈R时,f设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件(1) 当x∈R时,f(x-1)=f(-1-x)(2) f(1)=1(3) f(x)在R上的最小值为0.1 ,求f(x)的表达方
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/29 08:57:26
![设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件 (1) 当x∈R时,f设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件(1) 当x∈R时,f(x-1)=f(-1-x)(2) f(1)=1(3) f(x)在R上的最小值为0.1 ,求f(x)的表达方](/uploads/image/z/7624482-42-2.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dax2%2Bbx%2Bc%28a%2Cb%2Cc%E2%88%88R%2Ca%E2%89%A00%29%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6+%EF%BC%881%EF%BC%89+%E5%BD%93x%E2%88%88R%E6%97%B6%2Cf%E8%AE%BE%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dax2%2Bbx%2Bc%28a%2Cb%2Cc%E2%88%88R%2Ca%E2%89%A00%29%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6%EF%BC%881%EF%BC%89+%E5%BD%93x%E2%88%88R%E6%97%B6%2Cf%28x-1%29%3Df%28-1-x%29%EF%BC%882%EF%BC%89+f%EF%BC%881%EF%BC%89%3D1%EF%BC%883%EF%BC%89+f%28x%29%E5%9C%A8R%E4%B8%8A%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA0.1+%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A1%A8%E8%BE%BE%E6%96%B9)
设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件 (1) 当x∈R时,f设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件(1) 当x∈R时,f(x-1)=f(-1-x)(2) f(1)=1(3) f(x)在R上的最小值为0.1 ,求f(x)的表达方
设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件 (1) 当x∈R时,f
设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件
(1) 当x∈R时,f(x-1)=f(-1-x)
(2) f(1)=1
(3) f(x)在R上的最小值为0.
1 ,求f(x)的表达方式 2 ,求最大的m(m>1),使得存在t∈R,只要x∈[1,m],就有f(x+t)≤x恒成立
设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件 (1) 当x∈R时,f设二次函数f(x)=ax2+bx+c(a,b,c∈R,a≠0)满足条件(1) 当x∈R时,f(x-1)=f(-1-x)(2) f(1)=1(3) f(x)在R上的最小值为0.1 ,求f(x)的表达方
(1)∵当x∈R时,f(x-1)=f(-1-x),∴函数对称轴为x=-1
∴-b/2a=-1
a+b+c=1
f(-1)=0=a-b+c=0
∴a=1/4
b=1/2
c=1/4
f(x)=1/4 x^2+1/2 x+1/4
1 表达式为f(x)=1/4x^2+1/2x+1/4
2
f(x)=ax^2+bx+c(a,b,c∈R,a≠0),
由f(x-1)=f(-1-x)得-b/(2a)=-1,b=2a,
f(1)=a+b+c=3a+c=1,
4ac-b^2=4ac-4a^2=0,a=c=1/4,
∴f(x)=(1/4)(x+1)^2.
2,存在t∈R,只要x∈[1,m],就有f(x+t)≤x恒成立,
<==>(1/4)(x+t+...
全部展开
f(x)=ax^2+bx+c(a,b,c∈R,a≠0),
由f(x-1)=f(-1-x)得-b/(2a)=-1,b=2a,
f(1)=a+b+c=3a+c=1,
4ac-b^2=4ac-4a^2=0,a=c=1/4,
∴f(x)=(1/4)(x+1)^2.
2,存在t∈R,只要x∈[1,m],就有f(x+t)≤x恒成立,
<==>(1/4)(x+t+1)^2<=x,
<==>x^2+2(t+1)x+(t+1)^2<=4x,
<==>g(x)=x^2+2(t-1)x+(t+1)^2<=0,对x∈[1,m]恒成立,
<==>g(1)=1+2(t-1)+(t+1)^2=t^2+4t<=0,且
g(m)=m^2+2(t-1)m+(t+1)^2<=0,
<==>{t|-4<=t<=0}与A={t|t^2+(2m+2)t+m^2-2m+1<=0}的交集非空,
<==>0或-4∈A,
<==>m^2-2m+1<=0,或16-4(2m+2)+m^2-2m+1<=0,
<==>m=1,或m^2-10m+9<=0,
<==>1<=m<=9,
∴m的最大值=9.
收起