Mathematics
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OpenStudy (anonymous):
Please help me out. I will fan and medal
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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..
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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
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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..
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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?
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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?
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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..
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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 \[f^-1(x) = 12/2?\]
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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):
\(\dagger\)