証明f(x)=(3/4sin^4x+5) -1
由此,求f(x)的極大值和極小值.
証明果到:2【(cos^2)^2-(sin^2)^2 -2】/4sin^2+5
=2【(cos^2-sin^2)(cos^2+sin^2)-2】/4sin^2+5
=2【(cos^2-sin^2)(1) -2】/4sin^2+5
=2【(1-sin^2)-sin^2-2】/4sin^2+5
=2【-1-2sin^2】/4sin^2+5
=唔識做...
更新1:
點解要代sinx=0同sinx=1 ?
三角函數 中四
2012-04-07 7:53 pm
f(x)=2(cos^4x-sin^4x-2)/4sin^2x+5 <---個sin同cos既次方只係4,個x唔係次方尼
証明f(x)=(3/4sin^4x+5) -1
由此,求f(x)的極大值和極小值.
証明果到:2【(cos^2)^2-(sin^2)^2 -2】/4sin^2+5
=2【(cos^2-sin^2)(cos^2+sin^2)-2】/4sin^2+5
=2【(cos^2-sin^2)(1) -2】/4sin^2+5
=2【(1-sin^2)-sin^2-2】/4sin^2+5
=2【-1-2sin^2】/4sin^2+5
=唔識做...
証明f(x)=(3/4sin^4x+5) -1
由此,求f(x)的極大值和極小值.
証明果到:2【(cos^2)^2-(sin^2)^2 -2】/4sin^2+5
=2【(cos^2-sin^2)(cos^2+sin^2)-2】/4sin^2+5
=2【(cos^2-sin^2)(1) -2】/4sin^2+5
=2【(1-sin^2)-sin^2-2】/4sin^2+5
=2【-1-2sin^2】/4sin^2+5
=唔識做...
回答 (2)
2012-04-07 8:21 pm
✔ 最佳答案
f(x)= 2(cos^4x - sin^4x - 2)/(4sin^2x + 5)
= 2[(cos^2x - sin^2x)(cos^2x + sin^2x) - 2)]/(4sin^2x + 5)
= 2(-2sin^2x - 1)/(4sin^2x + 5)
= (-4sin^2x - 2)/(4sin^2x + 5)
= [-(4sin^2x + 5)+ 3]/(4sin^2x + 5)
= 3/(4sin^2x + 5) - 1
當sinx = 0 時﹐f(x)有極大值3/5 - 1 = 2/3
當sinx = 1時﹐f(x)有極小值3/9 - 1 = -2/3
2012-04-07 10:24 pm
在此題找maximum ,就要將 3/(4sin^2x + 5) - 1 裡的 (4sin^2x + 5) 變得最小.
找minimum,則要將(4sin^2x + 5) 變得最大.
∵0≦sin^2x≦1 (https://www.desmos.com/calculator/1i8cwloxae)
∴代sin^2x = 0就能得到這function的max.
同理, 代sin^2x = 1 就能得到這function的min.
2012-04-07 14:25:35 補充：
另外,也可以透過這方法找出max和min
0≦sin^2x≦1
0≦4sin^2x≦4
5≦4sin^2x+5≦9
1/9≦1/(4sin^2x+5)≦1/5
1/3≦3/(4sin^2x+5)≦3/5
-2/3≦3/(4sin^2x+5)-1≦-2/5
max = -2/5 , min = -2/3
找minimum,則要將(4sin^2x + 5) 變得最大.
∵0≦sin^2x≦1 (https://www.desmos.com/calculator/1i8cwloxae)
∴代sin^2x = 0就能得到這function的max.
同理, 代sin^2x = 1 就能得到這function的min.
2012-04-07 14:25:35 補充：
另外,也可以透過這方法找出max和min
0≦sin^2x≦1
0≦4sin^2x≦4
5≦4sin^2x+5≦9
1/9≦1/(4sin^2x+5)≦1/5
1/3≦3/(4sin^2x+5)≦3/5
-2/3≦3/(4sin^2x+5)-1≦-2/5
max = -2/5 , min = -2/3
收錄日期: 2021-04-26 19:17:27
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https://hk.answers.yahoo.com/question/index?qid=20120407000051KK00225