Reply With QuoteIS this is MGT101 (Financial Accounting)????? I dont think so..!!
MTH101 Assignment 3 Deadline 7 July 2010
Assignment 3View more random threads:
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Last edited by viki; 07-03-2010 at 08:24 AM.
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IS this is MGT101 (Financial Accounting)????? I dont think so..!!
its MTH 101 assignment # 3
i have made the correctionOriginally Posted by Taim
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[B]The more knowledge you have, the greater will be your fear of Allah.[/B]
Please Join My [B]Group Vuhelp[/B][B], Birthday Wishing, Daily Hadees[/B] [CODE][B]http://vuhelp.net/groups/vuhelp.html[/B]
[B]http://vuhelp.net/groups/birthday-wishing.html[/B]
[B]http://vuhelp.net/groups/daily-hadees.html[/B][/CODE]
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where is solution of mth101 assignment no 3 please Help
Brother is ka solution hay tu post karo please
INSHALLAH i will try
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[B]The more knowledge you have, the greater will be your fear of Allah.[/B]
Please Join My [B]Group Vuhelp[/B][B], Birthday Wishing, Daily Hadees[/B] [CODE][B]http://vuhelp.net/groups/vuhelp.html[/B]
[B]http://vuhelp.net/groups/birthday-wishing.html[/B]
[B]http://vuhelp.net/groups/daily-hadees.html[/B][/CODE]
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Best Answer – Q1
y = x^2+k\x = x^2 + kx^(-1)
Max/min points are found when the gradient = 0
Find the gradient:
dy/dx = 2x - kx^(-2)
2x - kx^(-2) = 0
Taking LCM
2x^3 - k =0
2x^3 = k
If x=3:
k = 2(3)^3 = 2(27) = 54
k = 54
Q2
f'(x) = 1 - 1/x^2 > or = 0 in [1, +oo]
So, f(x) is increasing.
fmin = f(1) = 1 + 1/1 = 2
x + 1/x > fmin = 2
f'(x) = 1 - 1/x^2 = (x^2 - 1)/x^2.
Then, if f(x) is increasing, then f'(x) > 0 and so:
(x^2 - 1)/x^2 > 0.
Then, since x^2 is positive for all x, we have:
x^2 - 1 > 0
==> x < -1 and x > 1.
Thus, f(x) is increasing for (1, infinity).
For the second part, since f(x) is increasing for x > 1:
x > 1 <==> f(x) > f(1).
Since f(1) = 1 + 1/1 = 2, we see that for x > 1:
f(x) > 1 + 1/1 = 2. ∴
I hope this helps!
Last edited by viki; 07-07-2010 at 02:42 PM.
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