函數(shù)y=2x-1的反函數(shù)______.
由y=2x-1得x=1+log2y且y>0
即:y=1+log2x,x>0
所以函數(shù)y=2x-1的反函數(shù)是y=1+log2x(x>0)
故答案為:y=1+log2x(x>0).
練習(xí)冊(cè)系列答案
相關(guān)習(xí)題

科目:高中數(shù)學(xué) 來(lái)源: 題型:

由函數(shù)y=f(x)確定數(shù)列{an},an=f(n),若函數(shù)y=f(x)的反函數(shù)y=f-1(x)能確定數(shù)列{bn},bn=f-1(n),則稱數(shù)列{bn}是數(shù)列{an}的“反數(shù)列”.
(1)若函數(shù)f(x)=2
x
確定數(shù)列{an}的反數(shù)列為{bn},求{bn}的通項(xiàng)公式;
(2)對(duì)(1)中{bn},不等式
1
bn+1
+
1
bn+2
+…+
1
b2n
1
2
loga(1-2a)
對(duì)任意的正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍;
(3)設(shè)cn=
1+(-1)λ
2
3n+
1-(-1)λ
2
•(2n-1)(λ為正整數(shù))
,若數(shù)列{cn}的反數(shù)列為{dn},{cn}與{dn}的公共項(xiàng)組成的數(shù)列為{tn},求數(shù)列{tn}前n項(xiàng)和Sn

查看答案和解析>>

科目:高中數(shù)學(xué) 來(lái)源: 題型:

(2007•浦東新區(qū)一模)由函數(shù)y=f(x)確定數(shù)列{an},an=f(n),若函數(shù)y=f(x)的反函數(shù)y=f-1(x)能確定數(shù)列{bn},bn=f-1(n),則稱數(shù)列{bn}是數(shù)列{an}的“反數(shù)列”.
(1)若函數(shù)f(x)=2
x
確定數(shù)列{an}的反數(shù)列為{bn},求bn;
(2)設(shè)cn=3n,數(shù)列{cn}與其反數(shù)列{dn}的公共項(xiàng)組成的數(shù)列為{tn}
(公共項(xiàng)tk=cp=dq,k、p、q為正整數(shù)).求數(shù)列{tn}前10項(xiàng)和S10
(3)對(duì)(1)中{bn},不等式
1
bn+1
+
1
bn+2
+…+
1
b2n
1
2
loga(1-2a)
對(duì)任意的正整數(shù)n恒成立,求實(shí)數(shù)a的范圍.

查看答案和解析>>

科目:高中數(shù)學(xué) 來(lái)源: 題型:

若函數(shù)y=f(x)存在反函數(shù)y=f-1(x),由函數(shù)y=f(x)確定數(shù)列{an},an=f(n),由函數(shù)y=f-1(x)確定數(shù)列{bn},bn=f-1(n),則稱數(shù)列{bn}是數(shù)列{an}的“反數(shù)列”.
(1)若數(shù)列{bn}是函數(shù)f(x)=
x+1
2
確定數(shù)列{an}的反數(shù)列,試求數(shù)列{bn}的前n項(xiàng)和Sn;
(2)若函數(shù)f(x)=2
x
確定數(shù)列{cn}的反數(shù)列為{dn},求{dn}的通項(xiàng)公式;
(3)對(duì)(2)題中的{dn},不等式
1
dn+1
+
1
dn+2
+…+
1
d2n
1
2
log(1-2a)對(duì)任意的正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍.

查看答案和解析>>

科目:高中數(shù)學(xué) 來(lái)源:浦東新區(qū)一模 題型:解答題

由函數(shù)y=f(x)確定數(shù)列{an},an=f(n),若函數(shù)y=f(x)的反函數(shù)y=f-1(x)能確定數(shù)列{bn},bn=f-1(n),則稱數(shù)列{bn}是數(shù)列{an}的“反數(shù)列”.
(1)若函數(shù)f(x)=2
x
確定數(shù)列{an}的反數(shù)列為{bn},求{bn}的通項(xiàng)公式;
(2)對(duì)(1)中{bn},不等式
1
bn+1
+
1
bn+2
+…+
1
b2n
1
2
loga(1-2a)
對(duì)任意的正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍;
(3)設(shè)cn=
1+(-1)λ
2
3n+
1-(-1)λ
2
•(2n-1)(λ為正整數(shù))
,若數(shù)列{cn}的反數(shù)列為{dn},{cn}與{dn}的公共項(xiàng)組成的數(shù)列為{tn},求數(shù)列{tn}前n項(xiàng)和Sn

查看答案和解析>>

同步練習(xí)冊(cè)答案