證明:
設(shè)Δx>0、x1>0,x2=x1(1+Δx),則x2>x1,f(1+Δx)>0
那么,f(x2)=f(x1(1+Δx))=f(x1)+f(1+Δx)
f(x2)-f(x1)=f(1+Δx)>0
∴ f(x)在(0,+∞)上是增函數(shù)
①在(0,+∞)上
∵f(4)=f(2×2)=f(2)+f(2)=2
又∵f(x)在(0,+∞)上是增函數(shù),f(2x-4)
已知函數(shù)f(x)(x∈R,x≠0)對(duì)任意的非零實(shí)數(shù)x1,x2,恒有f(x1x2)=f(x1)+f(x2)
已知函數(shù)f(x)(x∈R,x≠0)對(duì)任意的非零實(shí)數(shù)x1,x2,恒有f(x1x2)=f(x1)+f(x2)
當(dāng)x>1時(shí),f(x)>0,f(2)=1.
求證,f(x)在(0,+無(wú)窮)上是增函數(shù).
解不等式f(2x-4)
當(dāng)x>1時(shí),f(x)>0,f(2)=1.
求證,f(x)在(0,+無(wú)窮)上是增函數(shù).
解不等式f(2x-4)
數(shù)學(xué)人氣:855 ℃時(shí)間:2019-09-10 08:56:23
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