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已知函数F(x)=log2x+1x?1,g(x)=log2(x%1)(...

(1)f(x)在区间(1,+∞)上的单调递减.证明如下:任取1<x1<x2,则f(x1)-f(x2)=log2(x1+1)(x2?1)(x1?1)(x2+1)∵(x1+1)(x2?1)(x1?1)(x2+1)-1=2(x2?x1)(x1?1)(x2+1)∵1<x1<x2,∴2(x2?x1)(x1?1)(x2+1)>0∴(x1+1)(x2?1)(x1?1)(x2+1)>1∴f(...

(1)∵f(x)=log2(x+1),g(x)=log(3x+1)2,g(x)≥f(x),∴log2(x+1)≤log(3x+1)2,∴3x+1≥x+1>0,∴x≥0.(2)∵y=g(x)-f(x)=log(3x+1)2-log2(x+1)=log23x+1x+1(x≥0).令h(x)=3x+1x+1=3-2x+1,则h(x)为[0,+∞)上的增函数,∴h...

解,由条件可知函数定义域为(-1,1),f(-x)=x+log2(1+x/1-x).=-f(x)=x-log2(1-x/1+x).=x+log2(1+x/1-x)故函数是奇函数f(x)+f(-x)=0 (1)由上述论述可得f(1/2011)+f(-1/2011)=0 (2)存在最小值,将函数变形令g(x)=-x,易知函数在(...

解答: (1)2^x-1=0 ∴ 2^x=1 ∴ 2^x=2^0 ∴ x=0 满足x≤1 (2)1+log2(x)=0 ∴ log2(x)=-1 ∴ log2(x)=log2(1/2) ∴ x=1/2, 不满足x>1 综上,f(x)的零点是0

(1)设x<0,则-x>0,∵F(x)为R上的奇函数,∴F(x)=-F(-x)=-log2(4-x+1),∴x<0时,F(x)=-log2(4-x+1); (2)∵f(x)=log2(4x+1)+kx(k∈R)是偶函数,∴f(-x)=f(x)对任意x∈R恒成立,即log2(4-x+1)-kx=log2(4x+1)+kx恒成立...

(Ⅰ)当a=1时,f(x)=log2(x+1).∴f(x-1)=log2x,∴f(x)+f(x-1)=log2(x+1)+log2x=log2[x(x+1)],若f(x)+f(x-1)>0,则x>0x+1>0x(x+1)>1,解得:x∈(5?12,+∞),即x的取值范围为(5?12,+∞);(Ⅱ)∵函数g(x)是定义在R上奇...

(1)令X=x3,Y=y2,∴x=3X,y=2Y,∵点 (x,y) 是函数y=f (x) 图象上,∴2Y=log2(3X+1),即Y=12log2(3X+1),∴g (x)=12log2(3x+1)(x>-13);(2)由g(x)-f (x)≥0,得12log2(3x+1)-log2(x+1)≥0,∴3x+1>0x+1>03x+1≥(x+1)2,...

原题能拍出来吗?快

解:由题意可得g(x)=f(x)-log5|x-1|,根据周期性画出函数f(x)=(x-1)2的图象以及y=log5|x-1|的图象,根据y=log5|x-1|在(1,+∞)上单调递增函数,当x=6 时,log5|x-1|=1,∴当x>6时,y=log5|x-1|>1,此时与函数y=f(x)无交点.再根据y=...

(1)令x/3=x',,y/2=y'则x=3x',y=2y'代入f(x)=log2(x+1)得y'=12log2(3x'+1)则g(x)=1/2log2(3x+1) f(x)>g(x)即log2(x+1)>1/2log2(3x+1) ,所以(x+1)^2>3x+1>0且x+1>0解得x>1或-1/3

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