推 Indigoa :第三題上下同乘x-根號那坨 再取極限找吧~~~ 02/19 14:53
→ plc123 :非常謝謝你，我下次會改進 02/19 19:35
→ Rasin :這很明顯是要交差... 02/19 21:33
推 contentt :第三題 a=0 所以a/b=0 是嗎?! 02/20 15:57
※ 引述《plc123 (荷包蛋)》之銘言：
: 非常的抱歉我是用pdf檔的方式呈現，所以須先下載
: http://www.badongo.com/file/25110356
回答前兩題...
1. f(x) = (2^x+5)/(2^x-1)
反函數 整理
======> x = (2^y+5)/(2^y-1) ====> y = f^(-1)(x) = log [(x+5)/(x-1)]
2
=> [f^(-1)]'(x) = (1/ln2)(ln[(x+5)/(x-1)])' = -6/[(ln2)(x+5)(x-1)]
=> [f^(-1)]'(3) = -3/(8ln2)
2. x = 4 = t^3-2t => t = 2
dy/dt = 2(f(x)+3x)[(df/dx)(dx/dt)+3(dx/dt)]
= 2(f(t^3-2t)+3t^3-6t)[f'(x)(3t^2-2)+3(3t^2-2)]
(dy/dt)|t=2 = 2(f(4)+3*4)(f'(4)*10+3*10)
= 2*15*10(f'(4)+3) = 150
=> f'(4) + 3 = 1/2 => f'(4) = -5/2
剩下三題題目是: (下次要問問題有誠意點把題目打上來，便當狗很難用咧...)
3. lim [x+√(x^2+ax)] = b，b≠0，求b/a=?
x→∞
4. 如果 lim [(3-2kx)/(5+2kx)]^(4x) = 10，求k
x→∞
5. f(x) = x^3-|1-3x^3|，xε[-2,3]，求f在[-2,3]的最大值
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