函数f(x)=x-1/x+1,则f(x)+f(1/x)=?
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设y=x
(y-1)f[(y
1)/(y-1)]
f(y)=y
(1)
设y=(x
1)/(x-1)
[(y
1)/(y-1)-1]f{[(y
1)/(y-1)
1]/[(y
1)/(y-1)-1]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
[(y
1-y
1)/(y-1)f{[(y
1
y-1)/(y-1)]/[(y
1-y
1)/(y-1)]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
2/(y-1)f{[2y/(y-1)]/[2/(y-1)]})]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
2/(y-1)f(y)
f[(y
1)/(y-1)]=(y
1)/(y-1)
2f(y)
(y-1)f[(y
1)/(y-1)]=y
1
(2)
(2)-(1),得
f(y)=1
由y=x
,x≠-1(总感觉x≠-1这个条件不太对,应该是x≠1才对),得y≠-1
由y=(x
1)/(x-1)=1
2/(x-1),得y≠1
f(y)=1
(y≠±
1)
函数解析式为f(x)=1
(x≠±
1)
(y-1)f[(y
1)/(y-1)]
f(y)=y
(1)
设y=(x
1)/(x-1)
[(y
1)/(y-1)-1]f{[(y
1)/(y-1)
1]/[(y
1)/(y-1)-1]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
[(y
1-y
1)/(y-1)f{[(y
1
y-1)/(y-1)]/[(y
1-y
1)/(y-1)]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
2/(y-1)f{[2y/(y-1)]/[2/(y-1)]})]}
f[(y
1)/(y-1)]=(y
1)/(y-1)
2/(y-1)f(y)
f[(y
1)/(y-1)]=(y
1)/(y-1)
2f(y)
(y-1)f[(y
1)/(y-1)]=y
1
(2)
(2)-(1),得
f(y)=1
由y=x
,x≠-1(总感觉x≠-1这个条件不太对,应该是x≠1才对),得y≠-1
由y=(x
1)/(x-1)=1
2/(x-1),得y≠1
f(y)=1
(y≠±
1)
函数解析式为f(x)=1
(x≠±
1)
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