標題:
Solve x(lnx)=1.
此文章來自奇摩知識+如有不便請留言告知
發問:
我自己做: x(lnx)=1 diff. both sides w.r.t.x, lnx +1 =0 lnx=-1 x=1/e why putting x=1/e to the original equation is wrong? 更新: 抱歉,自己錯concept.以上做法只係搵出f '(x)=0 , 並非f (x)=0. f(x)=x(lnx)-1 ,for all x>0 f '(x)=lnx+1 ,for all x>0 f "(x)=1/x ,for all x>0 f '(x)=0 iff x=1/e f "(x) >0 ,for all x>0 更新 2: thus,for all x>0, f(x) attains its min. at x=1/e absolute min. value of f(x) = -(1+e)/e ,which is negative. lim(x→+無限) f(x)=+無限 i.e.f(x)=0 has at least one solution. Hence, please help.
最佳解答:
You can consider in this way: x ln x = 1 ln ( x^x ) = 1 ln ( x^x ) = ln e x^x = e [ not x = e here, remember ] Of course it has an unique solution for it is a continuous functions and we must have a root between 1.6 and 2, for 1.^1.6 e. However, there is no answer which can be expressed in Surd Form. What you can do is to work out a very close value by Bisection Method, a method which has been removed from HKCEE syllabus since 2006. You can use it to find an approximate value correct to any decimal places you like. Hope I can help you. 2011-02-27 17:05:45 補充： For sure Yuk should understand now how the answer can be obtained and why is the original tackling approach unable to find the solution.
其他解答:
Solve x(lnx)=1. lnx = 1/x If you familiarize with the graph of ln x and 1/x, then it is not difficult to see that there is only one solution. On the other hand, consider f(x) = x(lnx) - 1 f(1) = -1 and f(2) = 0.3862 By intermediate value theorem, there should be at least one root in the interval (0.5,2) Also, f'(x) = (lnx) + 1 which is strictly increasing for x > 1 Since f(1) = -1 and f(x) -> infinity when x -> infinity and f(x) is a continuous function. There should be at least one root by intermediate value theorem To find out the solution, you can use the bisection method or Newton's method.|||||1.763222834|||||Consider Newton's method
留言列表
