一道高一代数题:y=f(x)的定义域为(负无穷,-1)并(1,正无穷)且为奇函数,f(3)=1,当x>2时,f(x)>0;对于任意的x>0,y>0有:f(x+1)+f(y+1)=f(xy+1).证明:f(x)在(1,正无穷)内单调递增.
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![一道高一代数题:y=f(x)的定义域为(负无穷,-1)并(1,正无穷)且为奇函数,f(3)=1,当x>2时,f(x)>0;对于任意的x>0,y>0有:f(x+1)+f(y+1)=f(xy+1).证明:f(x)在(1,正无穷)内单调递增.](/uploads/image/z/7838486-62-6.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E4%B8%80%E4%BB%A3%E6%95%B0%E9%A2%98%EF%BC%9Ay%3Df%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%28%E8%B4%9F%E6%97%A0%E7%A9%B7%2C-1%29%E5%B9%B6%281%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%29%E4%B8%94%E4%B8%BA%E5%A5%87%E5%87%BD%E6%95%B0%2Cf%283%29%3D1%2C%E5%BD%93x%3E2%E6%97%B6%2Cf%28x%29%3E0%3B%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84x%3E0%2Cy%3E0%E6%9C%89%EF%BC%9Af%28x%2B1%29%2Bf%28y%2B1%29%3Df%28xy%2B1%29.%E8%AF%81%E6%98%8E%EF%BC%9Af%28x%29%E5%9C%A8%281%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%29%E5%86%85%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E.)
一道高一代数题:y=f(x)的定义域为(负无穷,-1)并(1,正无穷)且为奇函数,f(3)=1,当x>2时,f(x)>0;对于任意的x>0,y>0有:f(x+1)+f(y+1)=f(xy+1).证明:f(x)在(1,正无穷)内单调递增.
一道高一代数题:
y=f(x)的定义域为(负无穷,-1)并(1,正无穷)且为奇函数,f(3)=1,当x>2时,f(x)>0;对于任意的x>0,y>0有:f(x+1)+f(y+1)=f(xy+1).
证明:f(x)在(1,正无穷)内单调递增.
一道高一代数题:y=f(x)的定义域为(负无穷,-1)并(1,正无穷)且为奇函数,f(3)=1,当x>2时,f(x)>0;对于任意的x>0,y>0有:f(x+1)+f(y+1)=f(xy+1).证明:f(x)在(1,正无穷)内单调递增.
证明:由于:f(x+1)+f(y+1)=f(xy+1)
则有:f(xy+1)-f(y+1)=f(x+1)
任取x1,x2属于(1,正无穷),且x1>x2
则f(x1)-f(x2)
=f[(x1-1)+1]-f[(x2-1)+1]
=f[(x1-1)/(x2-1) +1]
由于:x1>x2>1
则有:x1-1>x2-1>0
故:(x1-1)/(x2-1) >1
则(x1-1)/(x2-1) +1>2
又x>2时,f(x)>0
则:f[(x1-1)/(x2-1) +1] >0
即对任意x1,x2属于(1,正无穷),
当x1>x2时,恒有f(x1)>f(x2)
故f(x)在(1,正无穷)内单调递增
证明:任取x1,x2∈(0,+∞)且x1>x2.
则x1+1,x2+1∈(1,+∞)
由题 f(x1+1)-f(x2+1)=f(x1/x2+1)
又x1>x2,∴x1/x2+1>2
则f(x1/x2+1)>0
∴f(x1+1)>f(x2+1)
∴f(x)在(1,+∞)内单调递增
证明 设y>x>1,u=x-1>0,v=(y-1)/(x-1)>1,uv=y-1>0,
由f(u+1)+f(v+1)=f(uv+1)得
f(x)+f(v+1)=f(y)
f(y)- f(x)=f(v+1)
由v>1得v+1>2, f(v+1)>0,故f(y)-f(x)>0, f(y)>f(x), f(x)在(1,+∞)内单调递增.