Question

(1)已知f(x)=(a、b、c是常數(shù))的反函數(shù)是f^-1(x)=,求a+b+c的值.

(2)設(shè)點P(-1,-2)既在函數(shù)f(x)=ax²+b(x≤0)的圖象上,又在f(x)的反函數(shù)的圖象上,求f^-1(x).

Answer 1

解：(1)設(shè)y=,解得x=,

即f^-1(x)=,

因此,,

由對應(yīng)項系數(shù)相等得a=3,b=5,c=-2,

∴a+b+c=6.

(2)點P(-1,-2)在f(x)=ax²+b上,則-2=a(-1)²+b,                   ①

又∵點P(-1,-2)在f^-1(x)上,

∴點(-2,-1)在f(x)上.

∴-1=a(-2)²+b.                                                                ②

由①②聯(lián)立,解得a=,b=-.

∴f(x)= x²-(x≤0).

∴f^-1(x)=-(x≥-).