Question

函数f(x)=1/πx+1/sinπx-1/π(1-x) x为【1/2~1)求f(1)使函数连续

Answer 1

简单分析一下，答案如图所示

Answer 2

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)