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Old Saturday, October 30, 2010
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||@sibgakhan :

Salam, but let me try to solve it, though i dont know much about pure maths.


f(x)= x^2 -x-1
g(x)= 3/x


now fog(x)= f { g(x) } = f ( 3/x )
replace x by 3/x in f(x)

fog(x) = (3/x)^2 - (3/x) - 1

= (9/x^2) -3/x -1

The above expression can be simplified easily , taking LCM as x^2,

= 9 -3x - x^2
------------
x^2


Domain of f (x) = { all real numbers }
Domain of g (x) = { all real numbers except 0 }
Domain of fog(x) = { all real numbers except 0}






Lim x^3 -5x^2 + 6x / ( x^2 -9)
x--->3

Solution :

Applying limit to every factor, we get

Lim x^3 - Lim 5x^2 + Lim 6x / ( x^2 -9)
x-->3 x-->3 x-->3


The limit Lim 6x / ( x^2 -9) is in (c /0) form, so requires L hospital rule
x-->3


so taking d/dx to 6x / ( x^2 -9) in denominator and nominator separately, we get,

6/ (2x-9)

now putting value an applying limit, we get

Lim x^3 - Lim 5x^2 + Lim 6/ (2x-9)
x-->3 x-->3 x-->3


= 3^3 - 5(3)^2 + 6 /( 2*3 -9)

= 27 - 45 -2
= -20



03.

7< ( 1-2x) ≤ 10

can be broken to two in equalities

=> 7< 1-2x ||||||||| 1- 2x ≤ 10

=> 7-1 < -2x ||||||||| => -2x ≤ 10-1

=> 6< -2x ||||||||| => -2x ≤ 9

=> -6 > 2x ||||||||| => 2x > -9

=> -3 > x ||||||||| => x > -4.5
which means between -∞ , -3 ||||||||| which means between -4.5 , ∞



when plotted together.

__-9/2------4-------3_____-2_____-1_____0____ 1_____2_____3______4___
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