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1、8.2解析式(精讲)(基础版)使用条件已知函数类型定数 待系法i:设出含有待定系数的解析式解题思路解题思路ii:将已知条件代入,建立方程(组),通过解方程(组)求出相应的 待定系数换元法换元法解析式使用条件l形如y=f(gG)的函数/t = g(x),求出X=Q(t),换元注意给新元t范围解题 (2) x=。(t)将代入表达式求出f(t)思路 (3)将t换成X得到fG)的解析式,要注意新元的取值范围配凑法解题 思路使用条件形如f(gGFG)(1)由已知条件f (g G) =F G)将F G)改写成关于g G)的表达式(2)以X替代gG),得fG)的解析式,同时注意给出X的范解方程组已知条件/(
2、X)与f ( )或/(T)等的式子解题构造出另外一个等式组成方程组,通过解方程组求出fG)。一般把思路一 X换成-X或X的倒数等考点一待定系数法考点二换元法解析式卜考点三解方程组法考点四配凑法考点一待定系数法求解析式【例1】(2022.全国.高三专题练习)(多选)已知函数/(x)是一次函数,满足x) = 9x+8,则/(x)的解析式可能为()B. f(x) = 3x-2A. /(x) = 3x+2C. /(x) = -3x+4D, /(a:) = -3x-4【答案】AD【解析】设由题意可知/(/(1) =攵(而+人)+匕=/x+M+/? = 9x + 8 ,2 = 9k = 3 k = -3所
3、以L 7 ,解得7 。或 .所以/(x) = 3x+2或x) = -3x4.故选:AD.始+ 6 = 8b = 2 b = -4【一隅三反】(2022全国高三专题练习)已知人幻是一次函数,且满足新九+1) xl) = 2x+17,则式1)=.【答案】9【解析】设/(x)=Qx+b(4R0),贝ij 3/U+1)2/(x1) = 3办+3a+3/72以+2。2/?=qx+5q+/?,即 以+5。一a 2,ci 2,_+。=2尤+17不论x为何值都成立,二1,解得I.7/(x) = 2x+7,从而得/(1)=9.Z? + 5q = 17,。= 7,故答案为:91. (2022全国高三专题练习)已知
4、/(x) = 4x + 9,且为一次函数,求尤)=【答案】2x+3或2尤9.【解析】因为力为一次函数,所以设力=+可丘0), 所以/%) = /(辰+ Z?)=攵(Ax + Z?) + /? =k2% + /(左+ 1),因为/f(x) = 4x + 9,所以公x + b(& + l) = 4x + 9恒成立,所以尸=4仅化+ 1) = 9,解得:k = -2 b = -9所以/(x) = 2x+3 或 /(%)=2x9,故答案为:2x+3或一2x9.3(2022全国高三专题练习)已知/(x)是一次函数,且满足3x+l)-2x-l) = 2x+17,求/(x)= 【答案】2x + 7【解析】因
5、为/(可是一次函数,设/(%) =方+跳0),因为3/(x+l) 2/(xl) = 2x+17,所以3a(x + l) + Z? 2q(x l) + b = 2x+17,a = 2/、L r,所以/ x) = 2x + 7,故答案为:2%+ 7. 01fa = 2整理可得以+5a +匕= 2x+17,所以,可得5。+。= 17考点二换元法求解析式【例2】(2022.全国.高三专题练习)若/( + l) = x 则/(x)的解析式为()A. 力=%21B. /(%) = (X_2)2C. /(x) = x2 -3x4-1D. /(x) = x2 -3x4-2【答案】D【解析】设,= 4 + 1,
6、则4 = r 1,则/=工一五=。一1)2-。-1)=产一3,+ 2,所以函数/(X)的解析式为,(x) = f3X+2 .故选:D.【一隅三反】(2022全国,高三专题练习)若函数满足/(3x + 2) = 9x + 8,则尤)的解析式是()A. /(x) = 9x + 8B. .f(x)=3x + 2C. /。)=-3%-4D. /(x)=3x + 2 或/。)=-3%-4【答案】B【解析】设1 = 3x + 2,x =彳,所以/)= 9x彳+ 8 = 3。-2)+8=3/ + 2所以/(x)=3x + 2 .故选:B.【一隅三反】1.(2022全国高三专题练习)已知函数/( + 2)=
7、% + 4五+ 5,则尤)的解析式为【答案】/(x) = x2+1(x2)【解析】令6 + 2 = r,则也2,且x = (2,所以/=(一2+4” 2)+ 5 =/+ 1, (?2)所以%)=炉+1(无之2),故答案为:/(x) = x2+1(x2).1. (2022全国高三专题练习)若函数/满足,(3x+2)= 9%+8,则x)=一.【答案】3x+2【解析】令3x+2 = Z,可得x = 彳,所以/。)= 9乂彳+ 8 = 3,+ 2,所以/(x) = 3x+2,故答案为:3x+2.2. (2022全国高三专题练习)已知/(x-1) = /,则y = /(x)的解析式为.【答案】/(x)
8、= x2 + 2x+1【解析】令x1 =入 则x =,+ l,/。)= + 1)2=/+ 2/ + 1,故答案为:/(x) = x2 + 2x + 1.3. (2022全国高三专题练习)已知函数/在定义域R上单调,且不(。,+8)时均有/(/(x) + 2x) = l,则/(-2)的值为()A. 3B. 1C. 0D. -1【答案】A【解析】根据题意,函数/(x)在定义域R上单调,且工(。,+8)时均有/(/(九) + 2x) = l,则/(x) + 2x 为常数,设/(x) + 2x = E,则/(x) = 2x + /,则有/) = 2/ + / = 1,解可得方=一1,则/(x)= -2
9、x l,故/(2) = 4-1 = 3;故选:A.考点三解方程组求解析式= x + 2,则/(2)=(【例3】(2022全国高三专题练习)若函数“X)满足A. 0A. 0B. 2C. 3D. -3【答案】D(i A(iAii (【解析】由x) 2/ =x + 2,可得:一2/(x) = + 2,联立两式可得司=一ax + - -2,代入.2可得/(2)= -3.故选:D.【一隅三反】1. (2022浙江高三专题练习)已知函数/W满足2/(x) + /(-x) = 3x,则火x)的解析式为()B . 7(x) = -3xD. /(x) = 3x2A. /(x) = 3xC. /(%) = 3f【
10、答案】A【解析】若/(x) = 3x,则 2/(x) + /(x) = 2x3x+(3x) = 3x,满足题意;若/(x) = 3以 则 2/(x) +/(X) = 2x(3x) + 3x = -3xw3x,不满足题意;若/(x)= 31,则 2/(x) + /(-x) = 2x(3f) + (3d) = 9fw3x,不满足题意;若/(x) = 3Y,plij 2/(x) + /(-x) = 2x 3x2 + 3x2 = 9x2 3x,不满足题意.故选:A.2.(2022.全国高三专题练习)已知定义域为的函数/(月满足2/3-/(-尤)=3%3,则/3=【答案】x31Q【解析】因为2x) r)
11、 = 3%3,所以2/(_x) “x) = 3%3,同除以2得/(t) 不# = 一不/,a Q两式相加可得不司=5V,即“x) = V.故答案为:p(143 (2022全国高三专题练习)若函数/G), g(x)满足/-2/ =2%,且x) + g(x) = x + 6,则 x7x/+ g(1)=.【答案】9【解析】由/(%)-27 -【解析】由/(%)-27 -41?=2/ ,可知 / _ 2/ (x) = 4x ,联立可得 /(x) = 2x,所以 /(1) = 2,/(-1) = -2 XfXX又因为/(l) + g(D = -1 + 6 = 5,所以 g(-l) = 5 + 2 = 7
12、,所以为(l) + g(-l) = 9.故答案为:9(1 A 24. (2022全国高三专题练习)已知3/(x) + 5/ = + 1,则函数/(x)的解析式为、X ) X3 S1【答案】/(%)=-二 OX oo【解析】3/(幻+ 5/ -16/(x)=1 Ox - 2,x21x + b.3/(? + 5/(x) = 2x + 1,x3-x5,得:二 /(1)=8x 813 51x +6故答案为:/(x) = -+ - + -o(SX oo考点四配凑法【例4】(2022全国高三专题练习)已知函数/(x-1) =/+2x-3,则/(x)=()A. x2+4xB. N+4C. x2+4x - 6
13、D. x2 - 4x - 1【答案】A【解析】/(x-1) = x2+2x-3 = (x-1)2+4(x-1),所以=+4%.故选:A【一隅三反】(2022浙江高三专题练习)已知了(九一 1) = %2一2%-3,贝U/(x)=.【答案】x2-4【解析】因为/(% l) = f 一2% 3 = (x l)24,所以/(x) = %24,故答案为:%2-4(2022全国高三专题练习)已知/。+ 3 = /+二,则/的值等于. XX【答案】7【解析】/(X + ) = + =(X + )2 2 , xX令/= x + L 当x0时,L.21 = 2,当且仅当 = 1时取等号, xV x当x0时,r = -(-x-) -2,当且仅当了 = -1时取等号, x/)=-2 ,, YO, -2 2,4W),/. /(x) = %2- 2 , xe(-oo, -2 U2, +oo)则/3) = 322 = 7故答案为:73.(2022.全国高三专题练习)已知xL)=N+J7,则/(x+_L)= XX-X【答案】丁十二+ 4X【解析】因为於一XX(1 、21+ 2 ,所以/(x) = X?+2 ,所以/(xH)= x-+ 2 = X2H7 + 4XX)X故答案为“+44.
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