若函数f(x)=ax/(x+1)在(2,+∞)上为增函数,求实数a的取值范围.我看到其中一种解法是f(x) = ax/(x+1) = (ax+a-a)/(x+1) = a - a/(x+1)在(2,+∞)上为增函数即a - a/(x+1)在(2,+∞)上单调增a/(x+1)在(2,+∞)上单
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![若函数f(x)=ax/(x+1)在(2,+∞)上为增函数,求实数a的取值范围.我看到其中一种解法是f(x) = ax/(x+1) = (ax+a-a)/(x+1) = a - a/(x+1)在(2,+∞)上为增函数即a - a/(x+1)在(2,+∞)上单调增a/(x+1)在(2,+∞)上单](/uploads/image/z/10266334-70-4.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%2F%28x%2B1%29%E5%9C%A8%282%2C%2B%E2%88%9E%29%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.%E6%88%91%E7%9C%8B%E5%88%B0%E5%85%B6%E4%B8%AD%E4%B8%80%E7%A7%8D%E8%A7%A3%E6%B3%95%E6%98%AFf%28x%29+%3D+ax%2F%28x%2B1%29+%3D+%28ax%2Ba-a%29%2F%28x%2B1%29+%3D+a+-+a%2F%28x%2B1%29%E5%9C%A8%EF%BC%882%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%E5%8D%B3a+-+a%2F%28x%2B1%29%E5%9C%A8%EF%BC%882%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E5%A2%9Ea%2F%28x%2B1%29%E5%9C%A8%EF%BC%882%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95)
若函数f(x)=ax/(x+1)在(2,+∞)上为增函数,求实数a的取值范围.我看到其中一种解法是f(x) = ax/(x+1) = (ax+a-a)/(x+1) = a - a/(x+1)在(2,+∞)上为增函数即a - a/(x+1)在(2,+∞)上单调增a/(x+1)在(2,+∞)上单
若函数f(x)=ax/(x+1)在(2,+∞)上为增函数,求实数a的取值范围.
我看到其中一种解法是
f(x) = ax/(x+1) = (ax+a-a)/(x+1) = a - a/(x+1)
在(2,+∞)上为增函数
即a - a/(x+1)在(2,+∞)上单调增
a/(x+1)在(2,+∞)上单调减
又因为在(2,+∞)上x+1单调增,1/(x+1)单调减
∴a>0
怎么能够由"a - a/(x+1)在(2,+∞)上单调增"得到"a/(x+1)在(2,+∞)上单调减"?
若函数f(x)=ax/(x+1)在(2,+∞)上为增函数,求实数a的取值范围.我看到其中一种解法是f(x) = ax/(x+1) = (ax+a-a)/(x+1) = a - a/(x+1)在(2,+∞)上为增函数即a - a/(x+1)在(2,+∞)上单调增a/(x+1)在(2,+∞)上单
你的做法很正确
常数a不影响单调性
a-a/(x+1)在(2,+∞)上单调增,即-a/(x+1)在(2,+∞)上单调增;
负号是改变单调性的,所以a/(x+1)在(2,+∞)上单调减
如果不懂,请Hi我,
a-a/(x+1)=(ax+a-a)/(x+1)=ax/(x+1)证明它为减函数就行了
是的啊
a - a/(x+1)在(2,+∞)上单调增
由于a是常数
所以
- a/(x+1)在(2,+∞)上单调增
即a/(x+1)在(2,+∞)上单调减那后面的"在(2,+∞)上x+1单调增"又是怎样得出的???a/(x+1)=a*1/(x+1) 看到x+1在(2,+∞)上x+1单调增 1/(x+1)在(2,+∞)上x+1单调减 而a/(x+1...
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是的啊
a - a/(x+1)在(2,+∞)上单调增
由于a是常数
所以
- a/(x+1)在(2,+∞)上单调增
即a/(x+1)在(2,+∞)上单调减
收起
re5