Re: [單變] 證反函數存在
※ 引述《woeichern (小暐 )》之銘言:
: 已知 f(x)= x^3 + x + 1 , 試證f(x)具有反含數 .
: 懇請高手不吝指教!
: 感謝!
→ woeichern:如果用非微積分方式該如何證明呢? 140.119.107.47 07/18 13:25
→ bulletproof:原po你是說不用微導去證明嗎? 218.169.66.246 07/18 13:26
→ zptdaniel:就想辦法把反函數求出來 123.194.96.150 07/18 13:27
→ zptdaniel:不過應該是求的你死去回來 123.194.96.150 07/18 13:27
→ zptdaniel:因為反函數存在不代表一定可以用我們 123.194.96.150 07/18 13:27
→ zptdaniel:所熟悉的式子寫出來 123.194.96.150 07/18 13:28
不用微積分一樣可以做這個問題 , 也不需要求出反函數
(i)
f 是 one-to-one
f(x) = f(y) => x^3 + x + 1 = y^3 + y + 1 => (x^3 - y^3) + (x-y) = 0
=> (x-y) * (x^2 + xy + y^2 + 1) = 0
因為 x^2 + y^2 ≧ |xy| , 所以 x^2 + xy + y^2 + 1 > 0
=> x = y
所以 f 是一個 injection
(ii)
f 是 onto
y 屬於 R , 考慮 x^3 + x + (1-y) = 0
=> 必存在一實數根 x0
=> f(x0) = y
所以 f 是一個 surjection
因此 , f 是一個 bijection , 故 f 的反函數存在
有錯請指教!
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