设函数f(x)=x^2-1,对任意x∈[2/3,+∞),f(x/m)-(4m^2)f(x)≤f(x-1)+4f(m)恒成立,则实数m的范围是这个方法不用考虑f(m)既然有那么m也应∈[2/3,+∞)吗把f(x)=x平方-1代入,得:x^2/m^2-1-4m^2(x^2-1)≤【(x-1)^2-1】+4(
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![设函数f(x)=x^2-1,对任意x∈[2/3,+∞),f(x/m)-(4m^2)f(x)≤f(x-1)+4f(m)恒成立,则实数m的范围是这个方法不用考虑f(m)既然有那么m也应∈[2/3,+∞)吗把f(x)=x平方-1代入,得:x^2/m^2-1-4m^2(x^2-1)≤【(x-1)^2-1】+4(](/uploads/image/z/3151581-69-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2-1%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E2%88%88%5B2%2F3%2C%2B%E2%88%9E%29%2Cf%28x%2Fm%29-%284m%5E2%29f%28x%29%E2%89%A4f%28x-1%29%2B4f%28m%29%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99%E5%AE%9E%E6%95%B0m%E7%9A%84%E8%8C%83%E5%9B%B4%E6%98%AF%E8%BF%99%E4%B8%AA%E6%96%B9%E6%B3%95%E4%B8%8D%E7%94%A8%E8%80%83%E8%99%91f%28m%29%E6%97%A2%E7%84%B6%E6%9C%89%E9%82%A3%E4%B9%88m%E4%B9%9F%E5%BA%94%E2%88%88%5B2%2F3%2C%2B%E2%88%9E%29%E5%90%97%E6%8A%8Af%28x%29%3Dx%E5%B9%B3%E6%96%B9-1%E4%BB%A3%E5%85%A5%2C%E5%BE%97%EF%BC%9Ax%5E2%2Fm%5E2-1-4m%5E2%28x%5E2-1%29%E2%89%A4%E3%80%90%EF%BC%88x-1%29%5E2-1%E3%80%91%2B4%EF%BC%88)
设函数f(x)=x^2-1,对任意x∈[2/3,+∞),f(x/m)-(4m^2)f(x)≤f(x-1)+4f(m)恒成立,则实数m的范围是这个方法不用考虑f(m)既然有那么m也应∈[2/3,+∞)吗把f(x)=x平方-1代入,得:x^2/m^2-1-4m^2(x^2-1)≤【(x-1)^2-1】+4(
设函数f(x)=x^2-1,对任意x∈[2/3,+∞),f(x/m)-(4m^2)f(x)≤f(x-1)+4f(m)恒成立,则实数m的范围是
这个方法不用考虑f(m)既然有那么m也应∈[2/3,+∞)吗
把f(x)=x平方-1代入,得:
x^2/m^2-1-4m^2(x^2-1)≤【(x-1)^2-1】+4(m^2-1)
展开,消去4m^2,得:x^2/m^2-1-4m^2x^2≤x^2-2x-4
把x^2项合并,常数合并,得:(1/m^2-4m^2-1)x^2≤-2x-3
因为x≠0,所以1/m^2-4m^2-1≤(-2x-3)/x^2
令y=(-2x-3)/x^2,x∈[3/2,+∞),对y求导,知当x在(-2,0)时y递减,在(-∞,-2】和【0,+∞)时递增.所以y的最小值在x=3/2处取到,此时y1=-8/3
所以1/m^2-4m^2-1≤-8/3.同乘m^2,整理得:12m^4-5m^2-3≥0
因式分解,(4m^2-3)(3m^2+1)≥0,所以4m^2-3≥0
即m∈(-∞,-根号3/2】∪【根号3/2,+∞)
设函数f(x)=x^2-1,对任意x∈[2/3,+∞),f(x/m)-(4m^2)f(x)≤f(x-1)+4f(m)恒成立,则实数m的范围是这个方法不用考虑f(m)既然有那么m也应∈[2/3,+∞)吗把f(x)=x平方-1代入,得:x^2/m^2-1-4m^2(x^2-1)≤【(x-1)^2-1】+4(
(x^2)/(m^2)-1-4m^2(x^2-1)≤x^2-2x+4(m^2-1)
[1/(m^2)-4m^2-1]x^2≤-2x-3
4m^2-1/(m^2)+1≥3*(1/x)^2+2*(1/x)
令g(1/x)=3*(1/x)^2+2*(1/x)
(1/x)∈(0,2/3]
4m^2-1/(m^2)+1≥g(1/x)max=g(2/3)=8/3
m^2≥3/4
m≤-√(3)/2或m≥√(3)/2