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Mathematics 7 Online
OpenStudy (alystar):

I get a little confused by these. For 𝑓(π‘₯) = π‘₯ βˆ’ 2, find 𝑓^βˆ’1(π‘₯). Then find 𝑓(𝑓^βˆ’1(8))

OpenStudy (alystar):

I know that 𝑓(𝑓^βˆ’1(8)) is 8

OpenStudy (alystar):

my answer is either 𝑓^βˆ’1(π‘₯) = βˆ’π‘₯ + 2 ; 8 or f^-1(x)=x+2;8

OpenStudy (alystar):

I'm just not sure which one

OpenStudy (alystar):

@Vocaloid could you perhaps clarify this for me?

myininaya (myininaya):

do you know how to solve y=x-2 for x?

myininaya (myininaya):

like to get x by itself?

OpenStudy (alystar):

yeah x=2+y

myininaya (myininaya):

cool now that is basically it interchange x and y and you have your inverse function

myininaya (myininaya):

y=2+x or you can write as y=x+2

myininaya (myininaya):

where you can replace y with f^(-1)(x)

OpenStudy (alystar):

Right. So would it be f^-1(x)=x+2;8?

myininaya (myininaya):

yes y=x-2 y+2=x so f^(-1)(x)=x+2

OpenStudy (alystar):

Awesome, thanks so much!

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