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Mathematics 14 Online
OpenStudy (howard-wolowitz):

Need help! Will medal!

OpenStudy (howard-wolowitz):

@Kawaii!!!2026

OpenStudy (anonymous):

ok do you know how to get rid of that fraction on the second one

OpenStudy (howard-wolowitz):

no

OpenStudy (anonymous):

ok see how it is divided by 2 all you have to do is times by 2

OpenStudy (anonymous):

then figure out the rest

OpenStudy (howard-wolowitz):

ok after that then what?

OpenStudy (howard-wolowitz):

ok i got x-1

OpenStudy (anonymous):

\(f\) and \(g\) are inverses of each other if \(f(g(x))=g(f(x))=x\).

OpenStudy (howard-wolowitz):

ok gotcha

OpenStudy (howard-wolowitz):

but how does that help me?

OpenStudy (howard-wolowitz):

I mean you sai they are so that would be B. but just because of that ^

OpenStudy (anonymous):

I didn't say they *are*, I said they would be if they satisfy the condition I posted above. Notice the "if".

OpenStudy (howard-wolowitz):

btw cool name and dathmaul is awesome

OpenStudy (howard-wolowitz):

hmm well what would I do to see if thay are?

OpenStudy (anonymous):

You have to evaluate \(f(g(x))\). Given that \(f(x)=-2x+1\) and \(g(x)=\dfrac{x-1}{2}\), you have \[f(x)=-2x+1~~\implies~~f(\color{red}{g(x)})=-2\color{red}{g(x)}+1=-2\left(\frac{x-1}{2}\right)+1\] So does \(f(g(x))=x\)?

OpenStudy (howard-wolowitz):

no its -x+2

OpenStudy (anonymous):

Right, which means the condition is NOT satisfied, so \(f\) and \(g\) can't be inverses.

OpenStudy (howard-wolowitz):

ok AWESOME!

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