11.Given f(x)=(x+2)^2 and g(x)= 1」1-x
a)Find f[g(x)] and g[f(x)]
b)Show that f[g(2)] ≠g[f(2)]
F.4 maths *b
2007-10-23 7:56 am
回答 (2)
2007-10-23 8:29 am
✔ 最佳答案
11.a)
f[g(x)]
= f[1/(1-x)]
= {[1/(1-x)]+2} ^2
= [(1+2-2x)/(1-x)] ^2
= [(3-2x)/(1-x)]^2
= [(3-2x)^2]/[(1-x)^2]
g[f(x)]
= g[(x+2)^2]
= 1/[1-(x+2)^2]
= 1/(1-x^2-4x-4)
= 1/(-x^2-4x-3)
b)
From (a),
f[g(2)]
= [(3-2x2)^2]/[(1-2)^2]
= (-1)^2 / (-1)^2
= 1/1
= 1
g[f(2)]
= 1/(-2^2-4x2-3)
= 1/(-4-8-3)
= -1/15
Therefore, f[g(2)] ≠g[f(2)]
If there is a mistake, please inform me!
參考: My Maths knowledge
2007-10-23 8:38 am
a.Let x be g(x)= ( 1」1-x)
f[g(x)] =((1」1-x)+2)^2
=((1+2-2x)」(1-x))^2
=((3-2x)」(1-x))^2
Let x be f(x)=(x+2)^2
g[f(x)] = 1」1-(x+2)^2
=1」(-1-x)^2
=1」(1+x)^2
b. f[g(2)] ≠g[f(2)]
Since ((3-2x)」(1-x))^2 ≠ 1」(1+x)^2
Therefore f[g(2)] ≠g[f(2)]
f[g(x)] =((1」1-x)+2)^2
=((1+2-2x)」(1-x))^2
=((3-2x)」(1-x))^2
Let x be f(x)=(x+2)^2
g[f(x)] = 1」1-(x+2)^2
=1」(-1-x)^2
=1」(1+x)^2
b. f[g(2)] ≠g[f(2)]
Since ((3-2x)」(1-x))^2 ≠ 1」(1+x)^2
Therefore f[g(2)] ≠g[f(2)]
收錄日期: 2021-04-13 14:06:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071022000051KK04761