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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
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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?
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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 \(f(g(x))=x\)?
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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.