MTH101 Assignment 3 Deadline 7 July 2010

[viki]

Assignment 3



  mgt101 assignment 3 vuhelp.jpg (Size: 50.4 KB / Downloads: 175)

[Taim]

IS this is MGT101 (Financial Accounting)????? I dont think so..!!

[Nazia Fatima]

its MTH 101 assignment # 3

[viki]

IS this is MGT101 (Financial Accounting)????? I dont think so..!!

i have made the correction

[gulammmmm]

where is solution of mth101 assignment no 3 please Help

[gulammmmm]

Brother is ka solution hay tu post karo please

[viki]

Brother is ka solution hay tu post karo please

INSHALLAH i will try

[gulammmmm]

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!