问题 解答题
设幂函数f(x)=(a-1)xk(a∈R,k∈Q)的图象经过点(
2
,2)

(1)求a,k的值;
(2)求函数y=f(x)+
1
f(x)
的最小值.
答案

(1)由题意,得a-1=1⇒a=2;    

将点(

2
,2)代入f(x)=xm(
2
)m=2
,所以m=2

(2)由(1)知,f(x)=x2,于是y=x2+

1
x2

x2+

1
x2
≥2(当且仅当x2=1时取等号),

即当x=±1时,函数y=x2+

1
x2
取得最小值为2.

写作题

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