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C3 functions

http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2008-June/paper.php#Q4
part c
i got for the inverse 1x−1\frac{1}{x} -1 which told me that it's a graph of 1x\dfrac{1}{x} translated on unit down

so i thought the range was f(x)<−1f(x)< -1

but it's not why is that?
[QUOTE=thegreatwhale;66469010]http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2008-June/paper.php#Q4
part c
i got for the inverse 1x−1\frac{1}{x} -1 which told me that it's a graph of 1x\dfrac{1}{x} translated on unit down

so i thought the range was f(x)<−1f(x)< -1

but it's not why is that?

Because it links into the domain of the original function. Do you agree that the range of f(x) is less than 1/4?
Original post by RDKGames
Because it links into the domain of the original function. Do you agree that the range of f(x) is less than 1/4?


i don't quite understand...
[QUOTE=thegreatwhale;66469846]i don't quite understand...

Sketch f(x) for X>3 and tell me the range that you find
Original post by RDKGames
Sketch f(x) for X>3 and tell me the range that you find


i've sketched it but i don't know how i can get the range from it....
[QUOTE=thegreatwhale;66470300]i've sketched it but i don't know how i can get the range from it....

Plug in X=3 and the range is everything less than that answer. Hence why you exclude the equal bit, JUST less than. So range is everything less than f(3)
Original post by RDKGames
Plug in X=3 and the range is everything less than that answer. Hence why you exclude the equal bit, JUST less than. So range is everything less than f(3)


so x<3 ???
[QUOTE=thegreatwhale;66470388]so x<3 ???

Are you getting confused between range and domain? Domain is what you are allowed to PLUG IN, range is what you get out as a result of plugging in the domain. So when you plug in X=3, you get 1/4. From your sketch you should see that as X increases from 3, the y value gets smaller and smaller thus the range is less than 1/4.

For the inverse, the domain the range of the original therefore the domain would be X<1/4 for the inverse.
Original post by RDKGames
Are you getting confused between range and domain? Domain is what you are allowed to PLUG IN, range is what you get out as a result of plugging in the domain. So when you plug in X=3, you get 1/4. From your sketch you should see that as X increases from 3, the y value gets smaller and smaller thus the range is less than 1/4.

For the inverse, the domain the range of the original therefore the domain would be X<1/4 for the inverse.


if i plug in 3 into 1x−1 \dfrac{1}{x} -1 i get −23-\dfrac{2}{3} not 0.25 >.>
[QUOTE=thegreatwhale;66470630]if i plug in 3 into 1x−1 \dfrac{1}{x} -1 i get −23-\dfrac{2}{3} not 0.25 >.>

I said into f(x), you are plugging it into f-1(X)
Original post by RDKGames
I said into f(x), you are plugging it into f-1(X)


oh shintezewizel that's why i got a weird answer.... ah thanks so f(x)>0.25 ??
ImageUploadedByStudent Room1468788955.160967.jpg

Hence answer to part c is between 0 and 1/4 for x

The 0 comes from the fact that the range for the original function is between 0 and 1/4; it never reaches 0.


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[QUOTE=thegreatwhale;66470806]oh shintezewizel that's why i got a weird answer.... ah thanks so f(x)>0.25 ??

No, less than that, observe your sketch for X>3 (assuming you got it right)
Original post by RDKGames
No, less than that, observe your sketch for X>3 (assuming you got it right)


tried looking at the graph for a while and the graph isn't very tall but goes on long for "infinity" but the range is the height so it's from 0 to 1/4 ???
Original post by thegreatwhale
tried looking at the graph for a while and the graph isn't very tall but goes on long for "infinity" but the range is the height so it's from 0 to 1/4 ???


Exactly. So that becomes the domain for its inverse then.
Original post by RDKGames
Exactly. So that becomes the domain for its inverse then.


thanks a bunch

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