互為反函數(shù)的函數(shù)圖象間的關(guān)系練習(xí)題及答案參考
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1.點(diǎn)(3,5)在函數(shù)y=ax+b的圖象上,又在其反函數(shù)的圖象上,則a、b的值分別為( )
A.a=-1,b=7 B.a=-1,b=-8 C.a=-1,b=8 D.以上都不對(duì)
答案：C
解析：由已知得點(diǎn)(3,5)和點(diǎn)(5,3)在直線y=ax+b上,
2.已知f(x)= 的圖象關(guān)于直線y=x對(duì)稱,則a的取值是( )
A.-1 B.1 C.-2 D.0
答案：A
解析：由y= ,得x= ,故f-1(x)= ,因函數(shù)f(x)的圖象關(guān)于直線y=x對(duì)稱,則f(x)與其反函數(shù)f-1(x)為同一函數(shù),易得a=-1.
3.如圖,設(shè)函數(shù)y=1- (-10),則函數(shù)y=f-1(x)的圖象是圖中的( )
答案：B
解析：易知點(diǎn)(- ,1- )在原函數(shù)圖象上,故(1- ,- )在其反函數(shù)圖象上,首先排除A、C,又1-,故點(diǎn)(1- ,- )在直線x= 左邊,排除D,選B.
4.(四川成都模擬)已知f(x)= ,且f-1(x-1)的圖象的對(duì)稱中心是(0,3),則a的值為( )
A. B.2 C. D.3
答案：B
解析：f-1(x)= ,f-1(x-1)= ,其對(duì)稱中心是(0,a+1),a+1=3 a=2.
5.已知函數(shù)f(x)存在反函數(shù),若點(diǎn)(a,b)在f(x)的圖象上,則下列各點(diǎn)中必在其反函數(shù)圖象上的點(diǎn)是( )
A.(f-1(b),b) B.(a,f-1(a))
C.(f(a),f-1(b)) D.(f-1(b),f(a))
答案：C
解析：由(a,b)在y=f(x)圖象上,則(b,a)在y=f-1(x)圖象上,且b=f(a),則a=f-1(b).
6.設(shè)函數(shù)y=f(x)與y=g(x)的圖象關(guān)于直線y=x對(duì)稱,且f(x)=(x-1)2(x1),則g(x)=____________.
答案：1- (x0)
解析：由已知得函數(shù)y=g(x)為y=f(x)的反函數(shù),由y=(x-1)2(x1),得x=1- ,故g(x)=f-1(x)=1- (x0).
7.試求函數(shù)y=1+2x-x2(x1)和它的反函數(shù)的圖象的交點(diǎn).
解：由y=1+2x-x2(x1)求得其反函數(shù)為y=1+ (x2),
由
得1+2x-x2=1+ (12),
即2x-x2= ,
則x(2-x)= ,x( )2= ,
2-x=0或x =1.
x=2或x2(2-x)=1.
由x2(2-x)=1,得2x2-x3=1,
即(x2-1)+(x2-x3)=0,
(x-1)(x+1-x2)=0,
x=1或x= .
∵12,
x=2或x=1或x= .
y=f(x)和y=f-1(x)的交點(diǎn)有3個(gè),分別是(1,2)、(2,1)、( , ).
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8.對(duì)于［0,1］上所有x的值,函數(shù)f(x)=x2與其反函數(shù)f-1(x)的相應(yīng)函數(shù)值一定成立的關(guān)系式為( )
A.f(x)f-1(x) B.f(x)f-1(x)
C.f(x)f-1(x) D.f(x)=f-1(x)
答案：B
解析：結(jié)合f(x)與f-1(x)圖象即得.
9.函數(shù)y=f(x)在［-1,2］上的圖象如圖所示,則f-1(x)x+1的解集為( )
A.［-1,0］ B.［0,1］
C.［-1,- ］ D.［-1,2］
答案：C
解析：由已知圖象易得f(x)=
故f-1(x)=
(1)當(dāng)01時(shí),f-1(x)x+1 x-1x+ 1x .
(2)當(dāng)-10時(shí),f-1(x)x+1 -2xx+1 x- ,-1- .由(1)(2)知所求解集為［-1,- ］.
10.點(diǎn)P在f(x)=1+ 的圖象上,又在其反函數(shù)的圖象上,則P點(diǎn)的坐標(biāo)為____________.
答案：(2,2)
解析：設(shè)點(diǎn)P的坐標(biāo)為(a,b),
由已知 解得
11.函數(shù)y= (x-1)的圖象與其反函數(shù)的圖象的交點(diǎn)坐標(biāo)為_________________.
答案：(0,0),(1,1)
解析：由y= ,得x= .
由x-1,得 1,即y2.
其反函數(shù)為f-1(x)= (x2).
由 得
12.已知函數(shù)f(x)= ,
(1)求反函數(shù)f-1(x);
(2)研究f-1(x)的單調(diào)性;
(3)在同一坐標(biāo)系中,畫出f(x)與f-1(x)的圖象.
解：(1)∵f(x)= =y,
x+5=y2,且y0.
x=y2-5.
則f(x)= 的.反函數(shù)為y=f-1(x)=x2-5(x0).
(2)由二次函數(shù)的圖象知當(dāng)x0時(shí),f-1(x)為增函數(shù).
(3)圖象如圖所示.
13.已知函數(shù)f(x)=2x2-4x+1,x［-1,0］.
(1)求f-1(x);
(2)作出y=f(x)和y=f-1(x)的圖象,并判斷其單調(diào)性;
(3)解不等式:f-1(7x)f-1(x+1).
解：(1)設(shè)y=2x2-4x+1=2(x-1)2-1,
2(x-1)2=y+1.
∵x［-1,0］,x-1［-2,-1］.
x-1=- .
f-1(x)=1- ,x［1,7］.
(2)y=f(x)和y=f-1(x)的圖象見圖.
∵y=f(x)在［-1,0］上是減函數(shù),
y=f-1(x)在［1,7］上是減函數(shù).
(3)由(2)知y=f-1(x)在［1,7］上是減函數(shù),
∵f-1(7x)f-1(x+1),
7x+11.解得 1,
即原不等式的解集為{x| 1}.
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14.已知函數(shù)f(x)的圖象過點(diǎn)(0,1),則f(4-x)的反函數(shù)的圖象過點(diǎn)( )
A.(1,4) B.(4,1) C.(3,0) D.(0,3)
答案：A
解析：∵f(x)的圖象過點(diǎn)(0,1),
f(0)=1,即f(4-4)=1.
f(4-x)圖象過點(diǎn)(4,1).
f(4-x)的反函數(shù)圖象過點(diǎn)(1,4).
15.設(shè)函數(shù)f(x)= ,已知函數(shù)y=g(x)的圖象與y=f-1(x+1)的圖象關(guān)于直線y=x對(duì)稱,則g(3)=______________.
答案：
解析：y=f-1(x+1) x+1=f(y) x=f(y)-1,
故y=f-1(x+1)的反函數(shù)為g(x)=f(x)-1= ,則g(3)= .
16.已知函數(shù)f(x)= .
(1)證明函數(shù)f(x)在定義域上有反函數(shù),并求出反函數(shù);
(2)反函數(shù)的圖象與直線y=x有無交點(diǎn)?
(1)證明:∵f(x)的定義域?yàn)檎龑?shí)數(shù)集,
當(dāng)0x2時(shí),f(x1)-f(x2)=( )-( )=( )(1+ )0.
f(x1)f(x2),即f(x)在(0,+)上是增函數(shù).
f(x)有反函數(shù).當(dāng)x(0,+)時(shí), (-,+).反函數(shù)的定義域?yàn)镽.
由y= ,得x-y -1=0.
解得 = .∵y ,
y- 0.而 0,
= ,x= (y+ )2.
f-1(x)= (x+ )2(xR).
(2)解：y=f-1(x)與y=f(x)的圖象關(guān)于y=x對(duì)稱,故只需判斷y=f(x)與y=x有無交點(diǎn).
由 得x= .
x(1- )=1.
當(dāng)01時(shí),01.0x(1- )1,此時(shí)方程無實(shí)數(shù)根.
當(dāng)x1時(shí),x(1- )0,方程無實(shí)根.
y=f(x)與y=x無交點(diǎn).
從而y=f-1(x)與y=x無交點(diǎn).
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