設f(x)為二次多項式,且f(x+1)-f(x)=2x,求f(x)?
然後第二題在下面
更新1:
補充一下第一題還有f(0)=2
式的運算請各位幫我 以下兩題?
回答 (1)
2019-07-25 3:06 pm
1
設f(x)為二次多項式,且f(x+1)－f(x)=2x，f(0)=2求f(x)?
Sol
設f(x)=ax^2+bx+c
f(x+1)=a(x+1)^2+b(x+1)+c
=a(x^2+2x+1)+b(x+1)+c
=ax^2+(2a+b)x+(a+b+c)
f(x+1)－f(x)
=2ax+a+b
2a=2，a+b=0
a=1，b=－1
f(0)=c=2
f(x)=x^2－x+2
2
(3x^3+2x^2+1)/(x+2)^3=a/(x+2)+b/(x+2)^2+c/(x+2)^3+d/(x+2)^4
Set x=－1
(－3+2+1)/1^3=a+b+c+d=0
(C)
設f(x)為二次多項式,且f(x+1)－f(x)=2x，f(0)=2求f(x)?
Sol
設f(x)=ax^2+bx+c
f(x+1)=a(x+1)^2+b(x+1)+c
=a(x^2+2x+1)+b(x+1)+c
=ax^2+(2a+b)x+(a+b+c)
f(x+1)－f(x)
=2ax+a+b
2a=2，a+b=0
a=1，b=－1
f(0)=c=2
f(x)=x^2－x+2
2
(3x^3+2x^2+1)/(x+2)^3=a/(x+2)+b/(x+2)^2+c/(x+2)^3+d/(x+2)^4
Set x=－1
(－3+2+1)/1^3=a+b+c+d=0
(C)
收錄日期: 2021-04-30 22:48:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190725062115AAliT8J