题目内容
解答下列问题:
(I)设f(x)=
(x≤-3),
(1)求f(x)的反函数f-1(x);
(2)若u1=1,un=-f-1(un-1),(n≥2),求un;
(3)若ak=
,k=1,2,3,…,求数列{an}的前n项和Sn.
(I)设f(x)=
x2-9 |
(1)求f(x)的反函数f-1(x);
(2)若u1=1,un=-f-1(un-1),(n≥2),求un;
(3)若ak=
1 |
uk+uk+1 |
(1)由y=f(x)=
(x≤-3),两边平方得出y2=x2-9,移向得,x2=y2+9
∵x≤-3,∴两边开方得出x=-
,(y≥0)
所以反函数为y=f-1(x)=-
(x≥0)
(2)由un=-f-1(un-1)得出un=-f-1(un-1)=
(n≥2),两边平方并移向得出un2-un-12=9
所以数列{un2}是公差为9的等差数列,且首项u12=1,
un2=1+(n-1)×9=9n-8,
∵un>0,∴un=
(3)ak=
=
(
-
),
x2-9 |
∵x≤-3,∴两边开方得出x=-
y2+9 |
所以反函数为y=f-1(x)=-
x2+9 |
(2)由un=-f-1(un-1)得出un=-f-1(un-1)=
un-12+9 |
所以数列{un2}是公差为9的等差数列,且首项u12=1,
un2=1+(n-1)×9=9n-8,
∵un>0,∴un=
9n-8 |
(3)ak=
1 | ||||
|
1 |
9 |
9k+1 |
9k-8 |
|
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