Mathematics
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OpenStudy (clongoria2):
Inverse function question. Help please!
Explain if possible!
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OpenStudy (clongoria2):
Which function below is the inverse of f(x) = 2x + 3/ 5
OpenStudy (clongoria2):
A. f^-1(x) = 5x + 3/2
B. f^-1(x) = 5x - 3/2
C. f^-1(x) = 5/2x + 3
D. f^-1(x) = 5/ 3 -2x
OpenStudy (anonymous):
Hello! Is it \[f(x)=\frac{ 2x+3 }{ 5 }\] or \[f(x)=2x+ \frac{ 3}{ 5 }\]?
OpenStudy (clongoria2):
Its the first one. Im having trouble understanding what to do after switching the x and y variables
OpenStudy (anonymous):
So you already switched x and y like this?
\[ x=\frac{ 2y+3 }{ 5 }\]
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OpenStudy (clongoria2):
Yes I have
OpenStudy (anonymous):
All right! So now multiply 5 on both sides.
OpenStudy (anonymous):
What would you get?
OpenStudy (clongoria2):
\[5x = 2y + 3\]
OpenStudy (anonymous):
Make the equation look like the slope-intercept form: \(y=mx+b\).
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OpenStudy (anonymous):
Do you know how?
OpenStudy (clongoria2):
Do you mean \[2y = 5x + 3\]
OpenStudy (clongoria2):
?
OpenStudy (anonymous):
Yes, but it should be \(2y = 5x - 3\). :)
Now divide both sides by 2, and that's it.
OpenStudy (clongoria2):
\[5x - 3 / 2 = y\]
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OpenStudy (anonymous):
\[y= \frac{ 5x-3 }{ 2 } \] or
\[y=\frac{ 5x }{ 2 }-\frac{ 3 }{ 2 }\]
OpenStudy (clongoria2):
The first one. Sorry.
OpenStudy (anonymous):
Oh, you can write it either way I put it, but it is not \(y=5x-\frac{ 3 }{ 2 }\).
OpenStudy (clongoria2):
I know i meant to do \[y = \frac{ 5x - 3 }{ 2}\]
OpenStudy (clongoria2):
Thank you so much!
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OpenStudy (anonymous):
You're welcome! :)