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Mathematics 14 Online
OpenStudy (clongoria2):

Inverse function question. Help please! Explain if possible!

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? \[ x=\frac{ 2y+3 }{ 5 }\]

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: \(y=mx+b\).

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):

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

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!

OpenStudy (anonymous):

You're welcome! :)

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