函数f(x)对任意m,n∈R,都有f(m+n=f(m)+f(n)-1,并且当x大于0,f(x)大于1(1)求证f(x)在R上是增函数(2)若f(3)=4,解不等式f(a^2+a-5)小于2
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![函数f(x)对任意m,n∈R,都有f(m+n=f(m)+f(n)-1,并且当x大于0,f(x)大于1(1)求证f(x)在R上是增函数(2)若f(3)=4,解不等式f(a^2+a-5)小于2](/uploads/image/z/9311484-12-4.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8Fm%2Cn%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28m%2Bn%3Df%28m%29%2Bf%28n%29-1%2C%E5%B9%B6%E4%B8%94%E5%BD%93x%E5%A4%A7%E4%BA%8E0%2Cf%28x%29%E5%A4%A7%E4%BA%8E1%281%29%E6%B1%82%E8%AF%81f%28x%29%E5%9C%A8R%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%282%29%E8%8B%A5f%283%29%3D4%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28a%5E2%2Ba-5%29%E5%B0%8F%E4%BA%8E2)
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函数f(x)对任意m,n∈R,都有f(m+n=f(m)+f(n)-1,并且当x大于0,f(x)大于1(1)求证f(x)在R上是增函数(2)若f(3)=4,解不等式f(a^2+a-5)小于2
函数f(x)对任意m,n∈R,都有f(m+n=f(m)+f(n)-1,并且当x大于0,f(x)大于1
(1)求证f(x)在R上是增函数
(2)若f(3)=4,解不等式f(a^2+a-5)小于2
函数f(x)对任意m,n∈R,都有f(m+n=f(m)+f(n)-1,并且当x大于0,f(x)大于1(1)求证f(x)在R上是增函数(2)若f(3)=4,解不等式f(a^2+a-5)小于2
1.∵f(m+n)=f(m)+f(n)-1
当m=n=0时,f(0)=f(0)+f(0)-1
∴f(0)=1
当m+n=0时,f(0)=f(m)+f(-m)-1
∴-f(m)=f(-m)-1
∴-f(x)=f(-x)-1
在R上任取x1>x2,则
f(x1)-f(x2)=f(x1)+f(-x2)-1
=f(x1-x2)-1
又∵当x>0时,f(x)>1
∴f(x1-x2)-1>0
∴f(x1)>f(x2)
因此该函数在定义域上单调递增
2.f(3)=f(1)+f(2)-1 =f(1)+f(1)+f(1)-1-1
∵f(3)=4
∴3f(1)-2=4
∴f(1)=2
∵f(a^2+a-5)