[微積] 幫忙求一微積課本題目~

看板Math作者 (假牙)時間14年前 (2011/06/14 09:02), 編輯推噓1(1推 0噓 10→)
留言11則, 2人參與, 最新討論串1/1
f(x)=(ax+b)/(cx+d) 假設bc-ad不等於0 1.求f^-1(x) (就是反函數QQ) 2.為何要具備bc-ab不等於0的條件? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.44.108.168
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06/14 09:05, , 1F
題目是(ax+b)/(cx+d)?
06/14 09:05, 1F
※ 編輯: Ru83u86 來自: 114.44.108.168 (06/14 09:08) ※ 編輯: Ru83u86 來自: 114.44.108.168 (06/14 09:08)
推
06/14 09:09, , 2F
當 bc-ad!=0, 設 x = (ay+b)/(cy+d)
06/14 09:09, 2F
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06/14 09:09, , 3F
=> (cx-a)y + (d-b) = 0 => y = (b-d)/(cx-a)
06/14 09:09, 3F
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06/14 09:09, , 4F
太帥了~感恩
06/14 09:09, 4F
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06/14 09:10, , 5F
若 bc-ad=0 則 f(x) 是常數 反函數不存在
06/14 09:10, 5F
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06/14 09:11, , 6F
f(x)=x^2+x+1 有反函數f^-1 求(f^-1)'(3)
06/14 09:11, 6F
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06/14 09:12, , 7F
(f^-1)(f(x)) = x => (f^-1)'(f(x))f'(x) = 1
06/14 09:12, 7F
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06/14 09:13, , 8F
(鏈鎖律.) 所以 x 代 1 得 (f^-1)'(f(1))f'(1) = 1
06/14 09:13, 8F
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06/14 09:13, , 9F
=> (f^-1)'(3) * 3 = 1 => (f^-1)'(3) = 1/3
06/14 09:13, 9F
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06/14 09:17, , 10F
或是 x 代 -2 得 (f^-1)'(3) = -1/3
06/14 09:17, 10F
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06/14 09:24, , 11F
...第三行是 b-dx 才對
06/14 09:24, 11F
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文章代碼(AID): #1DzhCIEq (Math)