函數(shù)解析式的習(xí)題兼答案

1、函數(shù)解析式的求法待定系數(shù)法：若已知函數(shù)的類型(如一次函數(shù)、二次函數(shù))，可用待定系數(shù)法;1.已知 f (x)是一次函數(shù)，且 ff (x) =x+2,則 f (x)=()A. x+1 B. 2x - 1 C. - x+1 D. x+1 或-x-1【解答】解：f (x)是一次函數(shù)，設(shè)f (x) =kx+b , ff (x) =x+2, 可得：k (kx+b) +b=x+2.即 k2x+kb+b=x+2, k2=1, kb+b=2.解得 k=1, b=1.則 f (x) =x+1 .故選：A.(2)換元法：已知復(fù)合函數(shù)f(g(x)的解析式，可用換元法，此時要注意新元的取值 范圍；9.若函數(shù) f (x)

2、滿足 f (3x+2) =9x+8,則 f (x)是()A. f (x) =9x+8 B. f (x) =3x+2C. f (x) =-3-4 D. f (x) =3x+2 或 f (x) = - 3x - 4【解答】解：令t=3x+2,貝U x=,所以f (t) =9三衛(wèi)+8=3t+2.33所以f (x) =3x+2.故選B .(3)配湊法：由已知條件f(g(x) = F(x), 以x怕t g(x),使得f(x)而解柝式;18.已知f (生!)=組;1則()工1A. f (x) =x2+1 (x冷)B. f (x)C. f (x) =x2- 1 (x聲)D. f (x)可將F(x)改寫成關(guān)于

3、g(x)的表達式，然后=x2+1 (x 聲)=x2- 1 (x 冷)【解答】解：由一 2出)也邙二區(qū)妝+1 一仁心)，得 f (x) =x2 1 ,又;2±1力，f (x) =x21 的 x力. 故選：C.19.已知 f (2x+1) =x2 2x-5,貝U f (x)的解析式為(A. f (x) =4x2 6 B. f (x)x _ 424C. f (x) =-x2+x- D. f (x) =x2-2x-5 424【解答】解：方法一：用 湊配法”求解析式，過程如下:f (2x+D =4 +1 ) 2+1)-學(xué)；424卡/、_ 1 2 315一或一丁方法二：用 換元法”求解析式，過程

4、如下：令 t=2x+1 ,所以，x= (t1), .f (t) =- (t- 1) 2-2是(t- 1) - 5=-t2-t- -42424f (x) =1x2- -x -,故選：B.424(4)消去法：已知f(x)與f或f( x)之間的關(guān)系式，可根據(jù)已知條件再構(gòu)造出另 外一個等式組成方程組，通過解方程組求出f(x).21.若 f (x)對任意實數(shù) x 恒有 f (x) -2f (-x) =2x+1,則 f (2)=(A. - B. 2 C. - D. 333【解答】解：:f (x)對任意實數(shù)x恒有f (x) - 2f (-x) =2x+1,用x代替式中的x可得f ( x) - 2f (x)

5、= - 2x+1,聯(lián)立可解得 f (x) =Zx - 1,f (2) =->2- 1=-1故選：C333函數(shù)解析式的求解及常用方法練習(xí)題.選擇題(共25小題)2 .若幕函數(shù)f (x)的圖象過點(2, 8),則f (3)的值為()A. 6 B. 9C. 16 D. 273 .已知指數(shù)函數(shù)圖象過點(1,夕，則f (-2)的值為()A. B. 4C. D. 2244 .已知f (x)是一次函數(shù)，且一次項系數(shù)為正數(shù)，若 ff (x) =4x+8,則f (x)=()A. 2x-p| B- - 2x-8 C. 2x-8 D. 2rpi或-2x 85 .已知函數(shù)f (x) =ax (a>0且aF

6、),若f (1) =2,則函數(shù)f (x)的解析式為( )A. f (x) =4x B, f (x) =2x C. f (z)二 * D. f =二)就42r i-k26.已知函數(shù) g二1一2萬，fg (x) -，則 f(0)等于()xA. - 3 B. -±C. -7 D. 322f2，.07.設(shè)函數(shù)f (x)='、，若存在唯一的x,滿足f (f (x) =8a2+2a,log* 工>0則正實數(shù)a的最小值是()A.3 B. 7 C. 7 D. 2S 428.已知f (x-1) =x2,則f (x)的表達式為()A. f (x) =x2+2x+1 B. f (x) =x2

7、-2x+1C. f (x) =x2+2x - 1 D . f (x) =x2 - 2x- 110 .已知 f (x)是奇函數(shù)，當 x>0 時£(K)二 - 4(1+，)，當 x<0 時 f (x)=()A.4(1-Q B. 4 g) C.尸7 (i+x) D.尸11 .已知 f (x) =lg (x 1),貝U f (x+3)=()A. 1g (x+1) B. 1g (x+2) C. 1g (x+3)D. 1g (x+4)12 .已知函數(shù) f (x)滿足 f (2x) =x，則 f (3)=()A. 0 B. 1 C. 1og23D. 313 .已知函數(shù)f (x+1) =

8、3x+2,貝U f (x)的解析式是()A. 3x-1 B, 3x+1C. 3x+2 D. 3x+414 .如果 f (工)一，則當x為且x為時，f (x)=()K 1 - XA. - B.C.D. - -1XX - 11 - IK15 .已知 f (")二 J/,則函數(shù) f (x)=()* '廣A. x B _! C組運 -2(x0)B,x22(x)C.x2 -2(|xp2)D. x2 216 .已知f (x- 1) =x2+6x,則f (x)的表達式是()A. x2+4x- 5B, x2+8x+7 C. x2+2x- 3 D. x2+6x- 1017 .若函數(shù)f (x)滿

9、足f 班+ 二出+1,則函數(shù)f0)的表達式是(A. x2 B, x2+1C. x2-2 D. x2- 120.若 f (x) =2x+3, g (x+2) =f (x-1),貝U g (x)的表達式為()A. g(x)=2x+1 B, g(x)=2x - 1C.g(x)=2x - 3 D. g(x)=2x+722.已知 f (x) +3f ( x) =2x+1 ,貝U f (x)的解析式是()A. f (x)=x+- B,f (x) = - 2x+-C. f (x)=-x+-D. f (x)= - x+-444223.已知f(x), g (x)分別是定義在R上的偶函數(shù)和奇函數(shù)，且f (x) -

10、g (x) =x15 B- 15 C-15 25.若f(x)滿足關(guān)系式f(x)出時秋則f(2)的值為(A. 1 B. - 1 C. - g D. W22二.解答題(共5小題)26.函數(shù) f (x) =m+1ogax (a>0且 a力)的圖象過點(8, 2)和(1, - 1). (I )求函數(shù)f (x)的解析式；(H)令g (x) =2f (x) - f (x - 1),求g (x)的最小值及取得最小值時 x的+x2+1,貝U f (1) +g (1)=()A. - 3 B. - 1 C. 1 D. 324.若函數(shù)f (x)滿足：f (x)-4fg)=x，則|f(x)|的最小值為(第7頁(

11、共8頁)W15-15-27 .已知f (x) =2x, g (x)是一次函數(shù)，并且點(2, 2)在函數(shù)fg (x)的圖 象上，點(2, 5)在函數(shù)gf (x)的圖象上，求g (x)的解析式.28 .已知 f (x) =, fg (x) =4-x, x+3(1)求g (x)的解析式；(2)求g (5)的值.29 .已知函數(shù) f (x) =x2+mx+n (m, nCR), f (0) =f (1),且方程 x=f (x)有兩個相等的實數(shù)根.(I)求函數(shù)f (x)的解析式；(H)當xq0, 3時，求函數(shù)f (x)的值域.30 .已知定義在R上的函數(shù)g (x) =f (x) x3,且g (x)為奇函

12、數(shù)(1)判斷函數(shù)f (x)的奇偶性；(2)若x>0時，f (x) =2x,求當x<0時，函數(shù)g (x)的解析式.函數(shù)解析式的求解及常用方法練習(xí)題參考答案與試題解析一.選擇題(共25小題)2 .【解答】解：幕函數(shù)f (x)的圖象過點(2, 8),可得8=2a,解得a=3,幕函 數(shù)的解析式為：f (x) =x3,可得f (3) =27.故選：D.3 .【解答】解：指數(shù)函數(shù)設(shè)為y=ax,圖象過點工)，可得：二a,函數(shù)的解22析式為：y=2 x,則 f (-2) =22=4.故選：B.4.【解答】解：設(shè)f (x) =ax+b, a>02 .f (f (x) =a (ax+b) +b=

13、ax+ab+b=4x+8,(2.一a = ” ,a=2f (x) =2x+-.故選：A.35 .【解答】解：=f (x) =ax (a>0, a力)，f (1) =2,f (1) =a1 =2,即 a=2,函數(shù)f (x)的解析式是f (x) =2x,故選：B.6 .【解答】解：令g (x) =1 - 2x=0則x=:則 f (- x) =-jrq (1 -x),又 f (x)是奇函數(shù)，所以 f (x) = - f ( - x) =q二 (1-x).故選 D.11 .【解答】解：f (x) =lg (x-1),則 f (x+3) =lg (x+2),故選：B.12 .【解答】解：函數(shù) f

14、(x)滿足 f (2x) =x,貝U f (3) =f ( 2103)=log23. 故選：C.13 .【解答】二葉(x+1) =3x+2=3 (x+1) - 1.f (x) =3x-1 故答案是：A14【解答】解：令工r，則x=-KtK 1 - X1f (t) =q,化簡得：f (t)= 即 f (x) =故選 B一工t _ 1x _ 1t15 .【解答】解：f (x+-)=JG= (xD) -2, f (x) =x2-2 (|xp2).故選：C.16 .【解答】解：f (x-1) =x2+6x,設(shè) x 1=t,貝U x=t+1 , .f (t) = (t+1) 2+6 (t+1) =t2+

15、8t+7,把t與x互換可得：f (x) =x2+8x+7.故選：B.17【解答】解：函數(shù)f (x)滿足f (4+=) =x+4+1 =函數(shù)f (x)的表達式是：f (x) =x2- 1. (x總).故選： 20【解答】解：用x- 1代換函數(shù)f (x) =2x+3中的x, 貝有 f (x 1) =2x+1 , . g (x+2) =2x+1=2 (x+2) - 3,g (x) =2x - 3,故選：C.22.【解答】解：vf (x) +3f (-x) =2x+1 , 用-x代替x,得：f ( x) +3f (x) = - 2x+1 ；-3X得：8f (x) =8x-2, - f (x) = -x

16、+f,故選：C.23【解答】解：由f (x) - g (x) =x3+x2+1,將所有x替換成-x,得 f(-x) - g( - x)=-x3+x2+1,根據(jù) f (x) =f ( - x), g ( - x) = - g (x),得f (x) +g (x) = X3+X2+1,再令 x=1,計算得, f (1) +g (1) =1.故選：C.24 .【解答】解：:f (x) -4f (1) =x, f(1)- 4f (x) =.1,聯(lián)立解得：f (x)=-(烏+Q ,15 K"x)T (由由)擊x g$， 故選B.當且僅當|x|=2時取等號,25 .【解答】解：=f (x)滿足關(guān)系

17、式f (x) +2f (1) =3x, f (2) +2f (=6, ®一 ,f)+2f(2)二1,乙乙一X2 得3f (2) =3, a f (2) =- 1,故選：B.二.解答題(共5小題)rf (Q)二工26【解答】解：(I)由，得、f (1) =-1nd-lo g,8=2nH-lo gQl= - 1解得m=-1, a=2,故函數(shù)解析式為f (x) = - 1+log2x,2(II) g(x) =2f(x) -f(x- 1) =2( - 1+log2x) - - 1+log2(x- 1)=l0g9-1 ,"一 其中x>1,因為當且僅當工- 1二一、即x=2時，=

18、"成立,I - 12而函數(shù) y=log2x - 1 在(0, +00)上單調(diào)遞增，貝U log2l>ln g24 - 1=1 , .L.故當x=2時，函數(shù)g (x)取得最小值1.27 .【解答】解：設(shè)g (x) =ax+b, a毛; 則：fg (x) =2ax+b, gf (x) =a?2x+b；丁根據(jù)已知條件有:22a+b=2、4a+b=5解得 a=2, b=-3;g (x) =2x-3.28 .【解答】解：(1)二.已知 f (x)=烏，fg (x) =4-x,x+3=1+x(X" 1).，二4 一工,且 g (x) ” 3.解得 g (x)M+3(2)由(1)可知：g (5) =123XS= -1.1+5229 .【解答】 解：(I) v f (x) =x2+mx+n,且 f (0) =f (1), n=1+m+n.(1 分)m= - 1. (2 分) . f (x) =x2-x+n .(3 分):方程x=f (x)有兩個相等的實數(shù)根，.二方程 x=x2 - x+n有兩個相等的實數(shù)根.即方程x2- 2x+n=0有兩個相等的實數(shù)根.(4分)( 2) 2 - 4n=0. (S 分) n=1.仁分). f (x) =x2 - x+1,17 分)(n)由(i)，知 f (x) =x2-x+1.此函數(shù)的圖象是開口