函數(shù)解析式的練習題兼答案精編版

1、最新資料推薦函數(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.換元法：已知復合函數(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)

2、f (x)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= L 2,所以 f (t) =9Xt 2+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柑t g(x),使得f(x)的解析式；可將F(x)改寫成關于g(x)的表達式，然后18.已知f (生!)=型工1A. f (x) =x2+1 (x加)C. f (x) =x2- 1 (x聲)則()B.

3、 f (x) =x2+1 (x 聲)D. f (x) =x2- 1 (x冷)10上 2 r【解答】解：由工+1)=2 工+1 1 +2-+1 _ =(k+1)2 _ 得 f (x) =x2- 1 ,又.工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)-與；424w_1 2 _ 315方法二

4、：用 換元法”求解析式，過程如下：令 t=2x+1,所以，x= (t1), f (t) =3 (t- 1) 2-2 (t- 1) - 5t2-芻-與42424 - f (x) =1x2-x ,故選：B.424(4)消去法：已知f(x)與f或f( x)之間的關系式，可根據(jù)已知條件再構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

5、) - 2f (x) =- 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題一.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 (a0且

6、aF),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 =(J) * D. f =(J)就42r i-k26.已知函數(shù) g二1 & , fg (x) -，則 f(0)等于()xA. - 3 B. -C. -7 D. 322f2，.07.設函數(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.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- 2x+1C. f (x

7、) =x2+2x - 1D . f (x) =x2 - 2x- 110 .已知f (x)是奇函數(shù)，當x0時(K)二(l+x),當x0且a力)的圖象過點(8, 2)和(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)年，fg (x) =4-x,(1)求g (x)的解析式；(2)求g (5)的值.29 .已知函數(shù) f

8、(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；(n)當xqo, 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)為奇函數(shù)(1)判斷函數(shù)f (x)的奇偶性；(2)若x0時，f (x) =2x,求當x02 .f (f (x) =a (ax+b) +b=ax+ab+b=4x+8,a 2 -24貝U f (0) =-=7=3故選 D 笠47.【解答】解：由f (f (x) =8a2+2a可化為2x=8a2+2a 或 10g2x=8a2+2a；則

9、由 0 2x 0;故解8a2+2a當?shù)茫琣g;故正實數(shù)a的最小值是故選B.28.【解答】解：二函數(shù)f (x- 1) =x .f (x) =f (x+1) 1= (x+1) 2=x2+2x+1故選 A.10.【解答】解：當x0,=4 ，Lab+b=8a=2f (x) =2x+-.故選：A .35 .【解答】解：=f (x) =ax (a0, 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=1則 f (- x) =-jrq (1 -x),又 f (x)是奇函數(shù)，所以 f (

10、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 (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：14 【解答】解：令工” 則x=l KtK 1 - X1- f (t)=7,化簡得：f (t)= 即f (x)故選B t-1s- 1t15 .【解答】解：f (x+-)=/二=(乂口)2 -2,- f (x) =

11、x2-2 (|x|或).故選:C.16 .【解答】解：:f (x-1) =x2+6x, 設 x 1=t,貝 x=t+1 , .f (t) = (t+1) 2+6 (t+1) =t2+8t+7,把1與乂互換可得：f (x) =x2+8x+7.故選：B.17【解答】解：函數(shù)f (x)滿足f (4力 =x+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：f

12、 (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+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+x而函數(shù) y=log2x - 1 在(0, +oo)上單調(diào)遞增，貝U log27T- 1=1 ,.L.故當x=2時，函數(shù)g (x)取得最小值1.27 【解答

13、】解：設g (x) =ax+b, a毛; 則：fg (x) =2ax+b, gf (x) =a?2x+b；+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 (!) -4f (x) =.1,聯(lián)立解得：f (x) =- (+Q,15 KX)T (由由)擊X%|$， 故選B.當且僅當|x|=2時取等號,25 .【解答】解：=f (x)滿足關系式f (x) +2f (工)=3x,f (2) +2f=6, 一 .f () +2f(2)喙1,乙乙一X2 得3f (2) =3, a f (2) =- 1,故選：B.二.解答

14、題(共5小題)F (226【解答】解：(I)由得iI 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(n ) g(x) =2f(x) -f(x- 1) =2( - 1+log2x) - - 1+log2(x- 1)=l0g9-1 ,其中x1,因為當且僅當工-1=3即x=2時，=成立,I - 1丁根據(jù)已知條件有:22a+b=2、4a+b=51 華彳等 a=2, b= 3; , g (x) =2x 3.28【解答】解：(“已知小)=黑，的(x)=4-x,=1+x(X 1).5?)口二4 一 工,且

15、g (x) ” 3.解得 g (x)M+3(2)由(1)可知：g (5)二二 61+5229.【解答】 解：(I) = 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.門分)(H)由(I),知f (x) =x2-x+1.此函數(shù)的圖象是開口向上，對稱軸為 工的拋物線.個分) 2當時，f (x)有最小值f (1).分)而 f 4)二(j)f(0)=1, f(3) =32-3+1=7.(門分)2224.當xC0, 3時,函數(shù)f (