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| 2017/06/21 06:53:16瀏覽44｜回應0｜推薦0 | |
| 標題: 該用什麼方法找 x?? 發問: 4^x +32=6^(x+2) 試找出x. 該用什麼方法最快找到答案﹖﹖ 最佳解答: 1. 最快? 用 Graphic Calculator la, 你有無? 無可以用 online free o既 GCalc2 @ http://gcalc.net/ 將 4^x + 32 同 6^(x+2), 兩條 function分開graph o係同一個graph, 相交點就係答案 我graph 過, 得到 x = -0.04960000005, y = 32.936.... 2. 唔淮graph 就唯有用 Newton's method / Linear Approximation, 搵 x, let f(x) = 4^x + 32 - 6^(x+2) = 0 then df/dx = f'(x) = (4^x)(ln4) - (6^(x+2))(ln6) Newton's method (only 1st degree approx shown here): x1 = x0 - [f(x=x0)/f'(x=x0)], let x0 = 0 (because we know the ans is close to 0) x1 = 0 - [4^0 + 32 - 6^(0+2) / (4^x)(ln4) - (6^(0+2))(ln6)] x1 = 0 - [ -3 / (ln4 - 36ln6)] x1 = -0.04753... plug -0.04753 入 4^x +32=6^(x+2) check 吓, 4^(-0.04753) + 32 = 32.936 6^(-0.04753 + 2) = 33.061 只係 1st degree approx 已經好近, 做埋 2nd 同 3rd degree, 一路做落去會越來越近 2nd degree x2 = x1 - [f'(x = x1) / f''(x = x1)] 3rd degree x3 = x2 - [f''(x = x2) / f''"(x = x2)], 如始類推... 其他解答: Use bisection method or Newton's method.|||||With numerical analysis like bisection method or Newton's method.
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