2e^(-x/2)(- x^2 + 5x - 4) + 1 = 0
更新1:
To : 自由自在 Please explain why you calculate f(9)? Why not f(10) or f(11) etc.?
Roots
2012-12-31 4:11 pm
Determine the number of real roots of the equation :
2e^(-x/2)(- x^2 + 5x - 4) + 1 = 0
2e^(-x/2)(- x^2 + 5x - 4) + 1 = 0
回答 (1)
2012-12-31 6:01 pm
✔ 最佳答案
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2012-12-31 11:35:07 補充：
It can be 9, 10, 11, etc. This is just to prove that there are values x>7 such that f(x)>0 so there must be a root in (7,x)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121231000051KK00052