函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/27 11:06:54
函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值
x){ھ iFqqFIɚv/|m3PBPS&֨xlc;74&qY-O$X;7{:mD8d $ Dm$ m#mUhZRG[T 0if"l Oyxo=m> Rjkd@`//+I5I5/.H̳ 

函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值
函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值

函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值
f(x+1)=[2^(x+1)][a(x+1)^2+b(x+1)+c]
=2^x[2ax^2+(4a+2b)x+2a+2b+2c]
f(x+1)-f(x)=2^x[ax^2+(4a+b)x+2a+2b+c]
∵f(x+1)-f(x)=2^x·x^2
∴2^x·x^2=2^x[ax^2+(4a+b)x+2a+2b+c]
即x^2=ax^2+(4a+b)x+2a+2b+c
比较系数得:a=1;4a+b=0;2a+2b+c=0;
解得:a=1,b=-4,c=6