作业帮设函数f(x)=(loga)x+2/x-2(a>0,且a≠1).x属于【m,n】是单调减函数,值域为【设函数f(x)=loga(x-2)/(x+2) x属于[m,n]是单调减函数,值域为[1+loga(n-1),1+log(m+1)]1.求a的范围2.求证n>4>m>2
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![设函数f(x)=(loga)x+2/x-2(a>0,且a≠1).x属于【m,n】是单调减函数,值域为【设函数f(x)=loga(x-2)/(x+2) x属于[m,n]是单调减函数,值域为[1+loga(n-1),1+log(m+1)]1.求a的范围2.求证n>4>m>2](/uploads/image/z/8981038-46-8.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%28loga%29x%2B2%2Fx-2%28a%26gt%3B0%2C%E4%B8%94a%E2%89%A01%29.x%E5%B1%9E%E4%BA%8E%E3%80%90m%2Cn%E3%80%91%E6%98%AF%E5%8D%95%E8%B0%83%E5%87%8F%E5%87%BD%E6%95%B0%2C%E5%80%BC%E5%9F%9F%E4%B8%BA%E3%80%90%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dloga%28x-2%29%2F%28x%2B2%29+x%E5%B1%9E%E4%BA%8E%5Bm%2Cn%5D%E6%98%AF%E5%8D%95%E8%B0%83%E5%87%8F%E5%87%BD%E6%95%B0%2C%E5%80%BC%E5%9F%9F%E4%B8%BA%5B1%2Bloga%28n-1%29%2C1%2Blog%28m%2B1%29%5D1.%E6%B1%82a%E7%9A%84%E8%8C%83%E5%9B%B42.%E6%B1%82%E8%AF%81n%3E4%3Em%3E2)
设函数f(x)=(loga)x+2/x-2(a>0,且a≠1).x属于【m,n】是单调减函数,值域为【设函数f(x)=loga(x-2)/(x+2) x属于[m,n]是单调减函数,值域为[1+loga(n-1),1+log(m+1)]1.求a的范围2.求证n>4>m>2
设函数f(x)=(loga)x+2/x-2(a>0,且a≠1).x属于【m,n】是单调减函数,值域为【
设函数f(x)=loga(x-2)/(x+2) x属于[m,n]是单调减函数,值域为[1+loga(n-1),1+log(m+1)]
1.求a的范围
2.求证n>4>m>2
设函数f(x)=(loga)x+2/x-2(a>0,且a≠1).x属于【m,n】是单调减函数,值域为【设函数f(x)=loga(x-2)/(x+2) x属于[m,n]是单调减函数,值域为[1+loga(n-1),1+log(m+1)]1.求a的范围2.求证n>4>m>2
令u=(x-2)/(x+2)=1-4/(x+2)
在(-∞,-2)和(2,+∞)上u单调递减,则一定有a>1.
根据定义域,m+1>0,则m>-1;
n-1>0,则n>1.
可见合适的m,n取值范围为n>m>2.
则:
f(m)=loga[1-4/(m+2)]=1+loga(m+1);
→1-4/(m+2)=a·(m+1);
m>2,则解得a>0.
a·m^2+(3a-1)m+2a+2=0.
f(n)=loga[1-4/(n+2)]=1+loga(n-1),
→1-4/(n+2)=a·(n-1)
an^2+(a-1)n-2a-2=0
∵1+loga(n-1)>1+log(m+1),又a>1,
∴n-1>m+1.
则n>m+2>4.
则由an^2+(a-1)n-2a-2=0得:
4^2·a+4(a-1)-2a-2>0.
a>1/3.