Urgent question (AS Maths paper1)

[Ritter]

I kindly need a fast reply as my exam is tomorrow.
My question is about functions.
suppose i'm given a quadratic function (2nd degree), and I'm told to find the inverse of the function.
When I solve it I get : ±√(x+...)

My question is how to determine if its + or - in the answer.
And how to determine if the domain/range is >=/<= or >/< only.

Thanks in advance.

 

[daviruss]

I kindly need a fast reply as my exam is tomorrow.
My question is about functions.
suppose i'm given a quadratic function (2nd degree), and I'm told to find the inverse of the function.
When I solve it I get : ±√(x+...)

My question is how to determine if its + or - in the answer.
And how to determine if the domain/range is >=/<= or >/< only.

Thanks in advance.

hehe i have the same problem bro i dnt knw how to put the range in which inequality signs

 

[mehdi1028]

For the given domain, if f(x) is decreasing function, keep the negative sign... if its increasing the keep the + sign.
For e.g n07_p1
q11. (v). Calculate g-1(x):
g(x)=2x^2 - 8x+11 for x=<2
g(x) = 2x^2 - 8x +11
g(x)=y
y= 2x^2 - 8x +11
make x the subject
2(x-2)^2 = y-3
(x-2) = ±√(.5(y-3)
For the given domain, g(x) is a decreasing function, therefore we keep the negative sign with the radical.
x= 2-√[(y-3)/2]
g(x)=y
g-1 ( y ) = 2 - √[(y-3)/2]
g-1(x) = 2 - √[(x-3)/2] Ans.

 

[deepum]

Hey..people..can anyone give all the difficult questions that could ever exist for p1....got my exams tomorrow..

 

[mehdi1028]

Hey..people..can anyone give all the difficult questions that could ever exist for p1....got my exams tomorrow..

Solve 2011 may June and Oct /Nov papers ... all three variants

 

[Anamol shrestha]

For the given domain, if f(x) is decreasing function, keep the negative sign... if its increasing the keep the + sign.
For e.g n07_p1
q11. (v). Calculate g-1(x):
g(x)=2x^2 - 8x+11 for x=<2
g(x) = 2x^2 - 8x +11
g(x)=y
y= 2x^2 - 8x +11
make x the subject
2(x-2)^2 = y-3
(x-2) = ±√(.5(y-3)
For the given domain, g(x) is a decreasing function, therefore we keep the negative sign with the radical.
x= 2-√[(y-3)/2]
g(x)=y
g-1 ( y ) = 2 - √[(y-3)/2]
g-1(x) = 2 - √[(x-3)/2] Ans.

..
in my opinion ..since domain of g(x) is less than or equals to 2 ( which is range of g-1) using - sign gives the value of g-1 less than 2..so - sign is used

 

[Ritter]

Thanks mehdi1028

But what about nov08 no.10 (iv)

function is: 6x - x^2 at x>=3
at this domain I think its a decreasing function, right?
so I should keep the - sign, but the mark scheme didn't.

And what about the domain/range when to put = sign in the equality and when to remove it?

 

[mehdi1028]

Thanks mehdi1028

But what about nov08 no.10 (iv)

function is: 6x - x^2 at x>=3
at this domain I think its a decreasing function, right?
so I should keep the - sign, but the mark scheme didn't.

And what about the domain/range when to put = sign in the equality and when to remove it?

h(x) is valid for x>=3
so h(x) is on the right side of the symmetry. Therefore only keep the positive sign with the radical.


I dont know abt the second part ... and dont worry abour range and domain .. that is of just 1 mark. You can still get 73 or 74

 

[anonymous123]

about the + or - sign...just remember that whenever the domain is less than the x-coordinate of the vertex then we put negative sign otherwise its positive

 

[Ritter]

h(x) is valid for x>=3
so h(x) is on the right side of the symmetry. Therefore only keep the positive sign with the radical.


I dont know abt the second part ... and dont worry abour range and domain .. that is of just 1 mark. You can still get 73 or 74

Sry but I don't understand. What do you mean by right side of symmetry. What's the difference between this and the example you mentioned. Could you please explain a little bit more?

And by the way it's out of 75 not 74

 

[Ritter]

Concerning the domain and range part:
part of ms, s12_11:

In the first column on the left where it's written: domain is x>k-4 or ...........
and on the third column it's written accept greater than represented in a bigger > sign.
The > sign on the left is smaller than that on the right. What's the difference between both of them?
and does the smaller mean >= or what?????