Re: [考古] 反函數
※ 引述《kusorz (^~^)》之銘言：
: 設f(x)=x+x^2+e^x且g(x)=f^-1(x).求g'(1)之值
-1
g = f
=> g(f(x)) = x
=> (g'(f(x)))(f'(x)) = 1
1
=> g'(f(x)) = -------
f'(x)
f(x) = x + x^2 + e^x
令 f(x) = 1 , 則 x + x^2 + e^x = 1 => x = 0
f'(x) = 1 + 2x + e^x
f'(0) = 1 + 2*0 + e^0 = 1 + 0 + 1 = 2
1 1
g'(1) = ------- = ---
f'(0) 2
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