Urgent Maths question

[cs]

Cie 9709 paper 01, November 2007

Q11 part (v), why in the last part, its - and not +-..? here's the question..

[studyorplay]

to avoid this confusion-you can just invert the completed square in part 11i) to get -->
1. 2(x-2)^2 +3 (completed square form)
2. x, -2 , ^2 , multiply by 2 , +3  (what the x value ungergoes in completed square form)
3.Inverse= x-3 divided by 2, then square root and finally add 2.
4. you will see that the domain has to be x>(or equal) 3
5.and from this you can deduce that the Range(y values) have to be >(or equal to) 2

to be honest im not really answering your question but this is just an alternative..

[Freaked12]

2(x-2)^2+3
Let g(x)=y     y=2(x-2)^2+3

We make x the subject

2(x-2)^2=y-3
(x-2)^2=(y-3)/2
(x-2)=+-SQR((y-3)/2)

For the given domain, g(x) is a decreasing function therefore we keep the negative
sign.+- Would make it a many to one function

so the inverse would be after substituting x with y
y=2-SQR((x-3)/2)

Now range of g inverse is the domain of g(x)
Domain of g(x) is X<=2
range of g inverse is g^-1(x)<=2

Note:For a One-one function g(x): Domain of g(x)=range of g inverse
and range of g(x)=domain of g inverse

[S.M.A.T]

This is because only only one to one function have inverse
many to one function do not have inverse.
The function g(x) is one to one function and g^-1(x) is the reflection of g(x) on line y=x,if you give +- together then g^-1(x) become many to one function which is not possible.Since g^-1(x) is the reflection of g(x) on line y=x it must also be one to one function

[cs]

Thanks studyorplay, Requiem(+rep) and asiftasfiq93.. !!