andyyeung0912召喚!

鋒bstc

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唔識計=-=

vbnn20

=.=咁似延數的,不過好似仲有幾課先幫到你...而且唔識英文....=.=好似講左好多廢話

andyyeung0927

數學我是愛莫能助我CE無A MATH, AL無任何MATH

angelliu

=.=咁似延數的,不過好似仲有幾課先幫到你...而且唔識英文....=.=好似講左好多廢話
vbnn20 發表於 2010-2-24 09:57 PM

果d p maths黎.
本帥打救你
a)
f'(x )=1/4x^(-3/4)-1/4 =1/4[1/(x^3/4) -1] f'(x ) = 0 for x=1 <0 for 1<x<inf. >0 for 0<x <1 therefore, f(x ) attains its max at x=1 for all x?(0,inf.) because f is cont. on [0,inf.) , => f(x ) <= f(1 ) x^1/4 - 1/4x -3/4 <=0
replace x by x/y >=0 (x/y)^1/4 <= 1/4(x/y)-3/4 x^1/4 * y^3/4 <= 1/4x -3/4y
b)
since F and G are non-negative, i.e. F(x ) >=0 and G(x )>=0 by a, sub x=F(x ) , Y= G(x ) [F(x )]^1/4 *[G(x )]^3/4 <= 1/4 F(x ) +3/4G(x ) in(a->b) [F(x )]^1/4 *[G(x )]^3/4 dx <= in(a->b) 1/4 F(x ) +3/4G(x ) dx =1/4*1+3/4*1 =1/4+3/4=1

極神

打到亂碼咁= =果然係神人...

angelliu

我不是神.....
我只是小小的天使;]

極神

呢題最難o既唔係條數,而係英文(對我來說)

angelliu

看來我的名字對你而言也不容易吧a"a!!

339122

果d p maths黎.
本帥打救你
a)
f'(x )=1/4x^(-3/4)-1/4 =1/4[1/(x^3/4) -1] f'(x ) = 0 for x=1
angelliu 發表於 2010-2-24 10:42 PM

哇 完全唔知你打緊d咩....

Rayc

呢d數對我來說真係深,看來我的數會愈來愈差
我連恆等式都未搞清楚