题目内容
求下列函数的表达式:
(1)一次函数f(x)使得f{f[f(x)]}=﹣8x+3,求f(x)的表达式;
(2)已知f(x)满足
,求f(x)的表达式.
(1)一次函数f(x)使得f{f[f(x)]}=﹣8x+3,求f(x)的表达式;
(2)已知f(x)满足
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046200922378.png)
解:(1)设f(x)=kx+b,k≠0,
则f[f(x)]=kf(x)+b=k2x+kb+b,
f{f[f(x)]}=k(k2x+kb+b)+b=k3x+k2b+kb+b=﹣8x+3,
∴
,解得k=﹣2,b=1,
∴f(x)=﹣2x+1.
(2)∵
,①
∴
,②
②×2,得
,③
③﹣①,得3f(x)=
,
∴
.
则f[f(x)]=kf(x)+b=k2x+kb+b,
f{f[f(x)]}=k(k2x+kb+b)+b=k3x+k2b+kb+b=﹣8x+3,
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046201202780.png)
∴f(x)=﹣2x+1.
(2)∵
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046201482488.png)
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046201762431.png)
②×2,得
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046202042550.png)
③﹣①,得3f(x)=
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046202311222.png)
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120904/201209042046202591762.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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