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Additional Math Help HERE ONLY...!

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BlackBunny103:
Hey guys, would you please help me with this question? It's from MJ 2007 P2.
Thanxx :D

adrian1993:
Blackbunny, I presume that you have to equate both equations?

Then use b square-4ac=>0.


I need help on May June 2003 Paper 2 Question 5 Part iii.

How do I get the domain and range of the inverse? I don't quite understand. Urgent help needed.

Ghost Of Highbury:

--- Quote from: BlackBunny103 on June 09, 2010, 03:59:10 pm ---Hey guys, would you please help me with this question? It's from MJ 2007 P2.
Thanxx :D

--- End quote ---

y = (x2 + 4)/2

y= 3x - k

(x2 + 4 ) /2 = 3x- k
(
k = -0.5x2 + 3x - 2

 -0.5x2 + 3x - 2 -k = 0

multiply both sides with -2

x^2 - 6x + (4+2k) = 0

now to intersect at a point, use b^2 - 4ac >= 0

36x^2 - 16x^2 - 8kx^2 >= 0

simplify

k =< 2.5

Ghost Of Highbury:

--- Quote from: adrian1993 on June 09, 2010, 04:09:24 pm ---Blackbunny, I presume that you have to equate both equations?

Then use b square-4ac=>0.


I need help on May June 2003 Paper 2 Question 5 Part iii.

How do I get the domain and range of the inverse? I don't quite understand. Urgent help needed.

--- End quote ---

The domain of f(x) is the range of f-1 (x)

The range of f(x) is the domain of f-1(x)

So the range of f-1(x) --> x >= 0 (domain of f(x)

and the domain of f-1(x) ---> x >= 0.5 (range of f(x)

range of f(X) = the least value of f(x) is when x = 0

when x = 0, f(x) = 0.5

so as x>=0 , f(x) >=0.5

elemis:
This stuff looks WAAAAAAY harder than my normal math.  :-\

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