close
標題:
Calculus Question
Given : (1) f(x) = 1 + 6e^(-kx) - 6.5e^(-2kx) where k is a +ve constant. (2) f(b) = f(b + 2.5) = 1.5 where b is a +ve constant. Find : (1) Maximum value of f(x). (2) Exact value of k. 更新: Note : b not equal to zero.
最佳解答:
1) Sub y = e-kk, we have: f(x) = 1 + 6y - 6.5y2 = 1 - 6.5 (y2 - 12y/13) = 1 - 6.5 (y2 - 12y/13 + 36/169) + 6.5 x 36/169 = 16/13 - 6.5 (y - 6/13)2 Hence max. value of f(x) occurs when y = 6/13 and the max. value is 16/13 2) With f(b) = f(b + 2.5) = 15: 1 + 6e-kb - 6.5e-2kb = 1 + 6e-k(b+2.5) - 6.5e-2k(b+2.5) 6e-kb - 6.5e-2kb = 6e-2.5ke-kb - 6.5e-5ke-2kb 6(1 - e-2.5k)e-kb = 6.5(1 - e-5k)e-2kb ekb = (13/12) (1 + e-2.5k) 1 + 6e-kb - 6.5e-2kb = 1.5 -0.5 + 6e-kb - 6.5e-2kb = 0 -0.5e2kb + 6ekb - 6.5 = 0 e2kb - 12ekb + 13 = 0 (ekb - 13)(ekb + 1) = 0 ekb = 13 or -1 (rejected) (13/12) (1 + e-2.5k) = 13 1 + e-2.5k = 12 e-2.5k = 11 -2.5k = ln 11 k = -(5/2) ln 11
其他解答:
Calculus Question
此文章來自奇摩知識+如有不便請留言告知
發問:Given : (1) f(x) = 1 + 6e^(-kx) - 6.5e^(-2kx) where k is a +ve constant. (2) f(b) = f(b + 2.5) = 1.5 where b is a +ve constant. Find : (1) Maximum value of f(x). (2) Exact value of k. 更新: Note : b not equal to zero.
最佳解答:
1) Sub y = e-kk, we have: f(x) = 1 + 6y - 6.5y2 = 1 - 6.5 (y2 - 12y/13) = 1 - 6.5 (y2 - 12y/13 + 36/169) + 6.5 x 36/169 = 16/13 - 6.5 (y - 6/13)2 Hence max. value of f(x) occurs when y = 6/13 and the max. value is 16/13 2) With f(b) = f(b + 2.5) = 15: 1 + 6e-kb - 6.5e-2kb = 1 + 6e-k(b+2.5) - 6.5e-2k(b+2.5) 6e-kb - 6.5e-2kb = 6e-2.5ke-kb - 6.5e-5ke-2kb 6(1 - e-2.5k)e-kb = 6.5(1 - e-5k)e-2kb ekb = (13/12) (1 + e-2.5k) 1 + 6e-kb - 6.5e-2kb = 1.5 -0.5 + 6e-kb - 6.5e-2kb = 0 -0.5e2kb + 6ekb - 6.5 = 0 e2kb - 12ekb + 13 = 0 (ekb - 13)(ekb + 1) = 0 ekb = 13 or -1 (rejected) (13/12) (1 + e-2.5k) = 13 1 + e-2.5k = 12 e-2.5k = 11 -2.5k = ln 11 k = -(5/2) ln 11
其他解答:
文章標籤
全站熱搜
留言列表
