已知函数f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)且f(x)在【0,1】上单调递减,则f(7/2)、f(7/3)、f(7/5)的大小?
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![已知函数f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)且f(x)在【0,1】上单调递减,则f(7/2)、f(7/3)、f(7/5)的大小?](/uploads/image/z/10345832-8-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0%E9%83%BD%E6%9C%89f%28-x%29%3Df%28x%29%2Cf%28x%29%3D-f%28x%2B1%29%E4%B8%94f%28x%29%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2C%E5%88%99f%287%2F2%29%E3%80%81f%287%2F3%29%E3%80%81f%287%2F5%29%E7%9A%84%E5%A4%A7%E5%B0%8F%3F)
已知函数f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)且f(x)在【0,1】上单调递减,则f(7/2)、f(7/3)、f(7/5)的大小?
已知函数f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)且f(x)在【0,1】上单调递减,
则f(7/2)、f(7/3)、f(7/5)的大小?
已知函数f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)且f(x)在【0,1】上单调递减,则f(7/2)、f(7/3)、f(7/5)的大小?
因为f[x]=-f[x+1]
所以 -f[x]=f[x+1]
f[-x]=f[x]
f[7/2]=f[5/2+1]=-f[5/2]=-f[3/2+1]=f[3/2]=f[1/2=1]=-f[1/2]=-f[-1/2+1]
=f[-1/2]=f[1/2]
f[7/3]=f[4/3+1]=-f[4/3]=-f[1/3+1]=f[1/3]
f[7/5]=f[2/5+1]=-f[2/5]=-f[-3/5+1]=f[-3/5]=f[3/5]
因为在【0,1】上递减
所以f[1/3]>f[1/2]>f[3/5]
ji f[7/3]>f[7/2]>f[7/5]
打这些累死了,吧分给我吧,嘿嘿~
由F(X)=F(-X)得偶函数,由F(X)=F(X+!)得周期为二,可以画出图像了,区间〔0,1〕的大小就很好比较了
f(7/2)=f(-7/2)=-f(-5/2)=f(-3/2)=-f(-1/2)=f(1/2)
f(7/3)=f(-7/3)=-f(-4/3)=f(-1/3)=f(1/3)
f(7/5)=f(-7/5)=-f(-2/5)=f(3/5)
3/5>1/2>1/3
f(7/5)