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# c3 functions watch

1. how do you find the inverse of f(x) = x+2 / x-5 and state its domain

thankkkkks

2. Change x to y and vice versa.

Rearrange to make y the subject.
3. Let and then rearrange to make x the subject (start by multiplying through by x-5). Then you have , so relabelling gives you .

To state its domain, think about the value(s) that the function can't take. Then the domain is everything except this/these value/values.
4. (Original post by Mr M)

Change x to y and vice versa.

Rearrange to make y the subject.
thankkks

the rearranging is the part i get stuck on..

my working out:

x = y+2 / y-5

x(y-5) = y+2

xy - 5x = y + 2

y = xy - 5x + 2

but surely this is wrong because there is still a y?
5. y-xy = y(1-x)
6. boromir (borrow me ear) has the correct approach.
7. (Original post by Mr M)
boromir (borrow me ear) has the correct approach.
Totally not fair dude, I can't do anything like that with your name, it's like you knew that well, what goes comes around
8. (Original post by boromir9111)
y-xy = y(1-x)
how do you get that?!

i can get the LHS but not the RHS? :/

thanks
9. How about multiplying both sides by (x-5) - that will remove the deonominator on the right hand side, then you get

y(x-5) = x + 2
yx - 5y = x + 2
take the x over to the other side, take the right over to the side the x came from and then
yx - x = 5y + 2
factorise
x(y-1) = 5y + 2
then divide by (y-1)
x = 5y+2 / y-1

so the inverse therefore would be 5x+2 / x-1.

The domain is the range of values that x can take. If you want to find that then look at the denominator on the inverse
10. (Original post by boomboompow.)
how do you get that?!

i can get the LHS but not the RHS? :/

thanks
expand out y(1-x) and you will see that it expands out to y-xy post here again if you don't understand that or anything else

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