答：
f(x)是以9為底數(shù)、9x+1為真數(shù)的對(duì)數(shù) 再加上 kx
一樣的問(wèn)題,定義域不是關(guān)于原點(diǎn)對(duì)稱(chēng)
請(qǐng)用文字表達(dá)一下f(x).偶函數(shù)或者奇函數(shù)一定關(guān)于原點(diǎn)對(duì)稱(chēng)的定義域
麻煩增加多一些括號(hào)?????????????????????η????????????????????????鷳????????9^x???9??x?η?....3^x???3??x?η?....f(x)=log9 [(9^x+1)+kx] ???????f(x)=log9(9^x+1)+kx????????????????д???f(x)??h(x)f(x)=log9(9^x+1)+kxh(x)=log9[a??3^x??-4/3a]??f(x)=log9 (9^x+1)+kx???????f(-x)=log9 [9^(-x)+1]-kx=f(x)=log9 (9^x+1)+kx?????log9 [(9^x+1)/9^x] -kx= log9 (9^x+1)+kxlog9 (9^x+1)-log9( 9^x) -kx=log9 (9^x+1?? +kx?????-x-kx=kx??????????(2k+1)x=0??????????2k+1=0????k=-1/2?????f(x)=log9(9^x+1)-x/2=log9(9^x+1)-log9[9^(x/2)]=log9(9^x+1)-log9 (3^x)=log9 [(9^x+1)/3^x]=log9 (3^x+1/3^x)>=log9(2)h(x)=log9 [a(3^x)-4a/3]h(x)=f(x)????3^x+1/3^x=a*3^x-4a/3??Ψ????(1-a)*(3^x)^2+(4a/3)*3^x+1=0??Ψ????g(t)=(1-a)t^2+(4a/3)t+1=0??t>01??a=1???????????(4/3)*t+1=0?????2??a>1???1-a<0????????g(t)???????￡??????t=(4a/3)/(2a-2)>0??g(0)=1>0?????????t>0???g(t)=0?????a>1???3????a<1???1-a>0????????g(t)??????????????t=(4a/3)/(2a-2)<0?????g(0)=1>0???????????t>0???g(t)=0??????????a>1