Re: [中學] 大學推甄題:反函數
※ 引述《agga (小孩)》之銘言：
: ※ 引述《t0444564 (艾利歐)》之銘言：
: : 少一句，[且若g(a)=b，則f(b)=a]
: 我覺得是大學推甄了, 所以下面這樣寫比較好
: 若 對所有定義域內的a, 滿足 g(f(a)=a, 且 f(g(a)=a
: 則稱g 為f的反函數
Let f:A->B
If there exists g:B->A s.t.
f(g(y))=y for all y in B and
g(f(x))=x for all x in A
then g could be denoted by f^-1
is the inverse of f and is unique.
eg1:
f(x)=x^2 ,which domain is R^+
g(y)=Sqrt(y) , which domain is R^+
f(g(y))=y for all y in R^+ and g(f(x))=x for all x in R^+
so g = f^-1
eg2:
f(x)=x^2 ,which domain is R^-
g(y)=-Sqrt(y) , which domain is R^+
f(g(y))=y for all y in R^+ and g(f(x))=x for all x in R^-
so g = f^-1
eg3:
f(x)=x^2 ,which domain is R
If g is the inverse of h,then g(f(x))=x for all x in R must hold.
But this will lead to the contradiction g(x^2)=x=g((-x)^2)=-x
So f has no inverse.
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