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Mathematics 15 Online
OpenStudy (anonymous):

Please help me out. I will fan and medal

OpenStudy (anonymous):

\[f^-1 (3)\] when \[f(x) = (2x+3)/5\]

OpenStudy (anonymous):

There?

OpenStudy (anonymous):

When I came here to help you out, you are already out.. :P

OpenStudy (anonymous):

Oh, you are here only..

OpenStudy (anonymous):

Let f(x) = y..

OpenStudy (anonymous):

\[y = \frac{2x+3}{5}\] Here, this is \(y\) in terms of \(x\),, can you find \(x\) in terms of \(y\) ?

OpenStudy (anonymous):

Im very sorry, my internet is really unstable :( But in this case, would x = 3 ? Since f^-1(3)?

OpenStudy (anonymous):

@eta

OpenStudy (anonymous):

What I got out of this is as follows: y = 2x+3/5 y = 2(3)+3/5 y = 6+3/5 y=9/5 Im not sure if this is right, though

OpenStudy (anonymous):

@eta

OpenStudy (anonymous):

See, you have to find \(f^{-1}(x)\) first..

OpenStudy (anonymous):

Do as I said above..

OpenStudy (anonymous):

@eta how would I find that though? I'm not sure how to

OpenStudy (anonymous):

you have: \[y = \frac{2x+3}{5}\]

OpenStudy (anonymous):

Multiply by \(5\) both the sides..

OpenStudy (anonymous):

What you get?

OpenStudy (anonymous):

\[5y = 10x + 15/25 ?\]

OpenStudy (anonymous):

Really?

OpenStudy (anonymous):

Can't you cancel 5 with denominator 5 on right side?

OpenStudy (anonymous):

\[5y = \frac{2x+3}{5} \times 5 \\ 5y = 2x + 3\] Isn't it?

OpenStudy (anonymous):

How can 5 cancel out if it is being multiplied by itself? Sorry, I am really bad at this :(

OpenStudy (anonymous):

One 5 is in numerator and other is in denominator, you can cancel it like this: \[\frac{2x+3}{\cancel{5}} \times \cancel{5} = 2x +3\]

OpenStudy (anonymous):

Ah, okay. I see now ^^

OpenStudy (anonymous):

What comes next, after \[5y = 2x +3?\]

OpenStudy (anonymous):

Now, subtract \(3\) from both the sides and tell me what you got?

OpenStudy (anonymous):

\[5y-3 = 2x?\]

OpenStudy (anonymous):

Good.. :) Now divide by \(2\) both the sides..

OpenStudy (anonymous):

\[5y-3/2 = x\]

OpenStudy (anonymous):

yep. So: \[x = \frac{5y-3}{2}\]

OpenStudy (anonymous):

Now, switch \(x\) and \(y\) to finally find your inverse. Replace x with y and y with x in the above equation..

OpenStudy (anonymous):

\[y = 5(x)-3/2\]. And x = 3, right?

OpenStudy (anonymous):

So, you will get: \[y = \frac{5x-3}{2}\]

OpenStudy (anonymous):

This is your f inverse, replace y here with \(f^{-1}(x)\). So: \[f^{-1}(x) = \frac{5x-3}{2}\]

OpenStudy (anonymous):

Here, put \(x=3\), yes right..

OpenStudy (anonymous):

Okay. So \[f^-1(x) = 12/2?\]

OpenStudy (anonymous):

And 12/2 simplifies to 6

OpenStudy (anonymous):

There you are..!! Good..

OpenStudy (anonymous):

So 6 is the final answer, yes? ^^

OpenStudy (anonymous):

Thank you very much ^^

OpenStudy (anonymous):

\(\dagger\)

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