F.4 Maths M1 differentiation

1. y= [(x-1)(x-2)] / [(x-3)(x-4)]
dy / dx = { [(x-3)(x-4)][(x-1)+(x-2)] - [(x-3)+(x-4)][(x-1)(x-2)] } / [(x-3)(x-4)]^2
= { [(x-3)(x-4)](2x-3) - (2x-7)[(x-1)(x-2)] } / [(x-3)(x-4)]^2
= [ (x^2-7x+12)(2x-3) - (2x-7)(x^2-3x+2) ] / [(x-3)(x-4)]^2
= [ (2x^3-17x^2+45x-36) - (2x^3-13x^2+25x-14) ] / [(x-3)(x-4)]^2
= (-4x^2+20x-22) / [(x-3)(x-4)]^2
= -4(x^2-5x+11/2) / [(x-3)(x-4)]^2
= -4{x^2-5x+[(5/2)^2-(5/2)^2]+11/2} / [(x-3)(x-4)]^2
= -4{(x-5/2)^2-(5/2)^2+11/2} / [(x-3)(x-4)]^2
= -4[(x-5/2)^2-3/4] / [(x-3)(x-4)]^2
when x>4, (x-5/2)^2 > (4-5/2)^2 = 9/4
=> (x-5/2)^2-3/4 > 3/2 => -4[(x-5/2)^2-3/4] < -6 < 0
As dy / dx < 0 ( [(x-3)(x-4)]^2 is positive and -4[(x-5/2)^2-3/4] is negative),
the function is decreasing when x>4.

2. Actually, we find maximum value and minimum value and they are usually local(or relative) maximum value and local minimum value. The local extremum(maximum or minimum) value at that point is compared with the values near that point. It is not necessary for local extremum be absolute extremum that means local maximum value can be less that local minimum value
eg. there is a function called f(x)
f(0.99) = 3.99
f(1) = 4
f(1.01) = 3.99
f(2.99) = 9.01
f(3) = 9
f(3.01) = 9.01
You see that f(1) is local maximum value as the value is relative maximum near x=1(eg. x=0.99 and x=1.01) and f(3)local minimum value as the value is relative minimum near x=1(eg. x=2.99 and x=3.01). However maximum value f(1) = 4 is less than minimum value f(3)=9


2014-06-18 11:26:48 補充：
sorry for inconvenience of reading the answer of Q2
the better version is posed in area of opinion

2014-06-18 11:29:32 補充：
Actually, we find maximum value and minimum value
and they are usually local(or relative) maximum value
and local minimum value.
The local extremum(maximum or minimum) value at that point
is compared with the values near that point.

2014-06-18 11:30:32 補充：
It is not necessary for local extremum be absolute extremum
that means local maximum value can be less than
local minimum value

2014-06-18 11:30:55 補充：
eg. there is a function called f(x)
f(0.99) = 3.99
f(1) = 4
f(1.01) = 3.99
f(2.99) = 9.01
f(3) = 9
f(3.01) = 9.01

2014-06-18 11:31:42 補充：
You see that f(1) is local maximum value
as the value is relative maximum near x=1
(eg. x=0.99 and x=1.01)
and f(3)local minimum value
as the value is relative minimum near x=1
(eg. x=2.99 and x=3.01).
However maximum value f(1) = 4 is less than minimum value f(3)=9

2014-06-21 09:00:48 補充：
there is a typo
f(3)local minimum value
as the value is relative minimum near x= "3"
(eg. x=2.99 and x=3.01).
However maximum value f(1) = 4 is less than minimum value f(3)=9

1.
y= [(x-1)(x-2)] / [(x-3)(x-4)]
In y=In (x-1) +In (x-2) - In (x-3) -In (x-4)
(dy/dx)(1/y)=1/(x-1) + 1/(x-2) - 1/(x-3) -1/(x-4)
dy/dx=y/(x-1) + y/(x-2) - y/(x-3) - y/(x-4)

2.
Because some functions' image maximum value at negative number, minimum value at positive number. So some functions' maximum value is less than minimum value.

2014-06-18 22:45:07 補充：
1.
x x<1 x=1=2=3=4 x>4

dy/dx + math error -

So the function is decreasing.