標題:
F.4 maths *b
發問:
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)]
最佳解答:
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!
其他解答:
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)]
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