數學 指數函數 急！！?
回答 (1)
2017-05-21 7:53 pm
(a)
f(x+d)-f(x)
=[300-280(0.5)^(x+d)]-[300-280(0.5)^x]
=280[(0.5)^x-(0.5)^(x+d)]
=280(0.5)^x (1-0.5^d)>0 (因為d>0, 0.5^d<1)
(b)
由(a)
x=0 時, 有最低溫度
f(0)=300-280(0.5)^0=300-280=20 (度C) ...Ans
(c)
f(x+2)-f(x)
=280(0.5)^x (1-0.5^2)
=280(0.5)^x (0.75)
=105
0.5^x=105/280/0.75=0.5
x=1 .....Ans:
(d)
f(x)=300-280(0.5)^x<300 (因為 280(0.5)^x>0)
f(x+d)-f(x)
=[300-280(0.5)^(x+d)]-[300-280(0.5)^x]
=280[(0.5)^x-(0.5)^(x+d)]
=280(0.5)^x (1-0.5^d)>0 (因為d>0, 0.5^d<1)
(b)
由(a)
x=0 時, 有最低溫度
f(0)=300-280(0.5)^0=300-280=20 (度C) ...Ans
(c)
f(x+2)-f(x)
=280(0.5)^x (1-0.5^2)
=280(0.5)^x (0.75)
=105
0.5^x=105/280/0.75=0.5
x=1 .....Ans:
(d)
f(x)=300-280(0.5)^x<300 (因為 280(0.5)^x>0)
收錄日期: 2021-04-24 00:26:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170521104511AAAfgsX