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Mathematics 21 Online
OpenStudy (anonymous):

The function of y=(f)x has an inverse function. The point (3,5) is on the graph of the function y=(f)x. What on the graph would be the inverse function?

OpenStudy (anonymous):

i guess that would be (5,3)

OpenStudy (anonymous):

since the inverse function switches the ordered pairs

OpenStudy (anonymous):

btw it is \[y=f(x)\] and \[y=f^{-1}(x)\] not for example \[y=(f)x\]

OpenStudy (anonymous):

in other words if \[f(3)=5\] then \[f^{-1}(5)=3\]

OpenStudy (anonymous):

I thought it would make them negative.. but if you're sure.. then I guess you're right.

OpenStudy (anonymous):

Oh, thank you.(:

OpenStudy (anonymous):

oh no i am sure. think of elementary algebra, and your job is to solve for x

OpenStudy (anonymous):

say \[2x+1=7\] you would say \[2x=6\] \[x=3\]

OpenStudy (anonymous):

right...

OpenStudy (anonymous):

That sounds right..

OpenStudy (anonymous):

what i meant was on the graph of \[y=2x+1\] is the point (3,7) plug in 3, get out 7

OpenStudy (anonymous):

what is the inverse? well you just found it. subtract 1, divide by 2 so your inverse is \[f^{-1}(x)=\frac{x-1}{2}\] plug in 7, get out 3

OpenStudy (anonymous):

on the graph of \[f(x)=2x+1\] is the point (3,7) and on the graph of \[f^{-1}(x)=\frac{x-1}{2}\] is the point (7,3) it switches them

OpenStudy (anonymous):

Oh, alright! I think I kinda get it, still really confusing. Just doing summer work, will be learning this in a few days none the less.

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