Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
f of x equals four divided by x. and g of x equals four divided by x
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OpenStudy (kellyspeakslouder):
@triciaal
OpenStudy (kellyspeakslouder):
@jim_thompson5910 this is a writing portion
satellite73 (satellite73):
\[f(x)=\frac{4}{x}\] right?
OpenStudy (kellyspeakslouder):
yes
satellite73 (satellite73):
and also \(g(x)=\frac{4}{x}\)
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OpenStudy (kellyspeakslouder):
yes
satellite73 (satellite73):
so what you need is to compute \[f(g(x))\] which only takes two steps
first \[f(g(x))=f(\frac{4}{x})\]
satellite73 (satellite73):
then since \[f(\heartsuit)=\frac{4}{\heartsuit}\] you get \[\large f(\frac{4}{x})=\frac{4}{\frac{4}{x}}\]
satellite73 (satellite73):
a little algebra will turn \[\frac{4}{\frac{4}{x}}\] in to \(x\), which is what you want
OpenStudy (kellyspeakslouder):
what are the hearts for?
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satellite73 (satellite73):
place holders so you can understand why \[\large f(\frac{4}{x})=\frac{4}{\frac{4}{x}}\]
OpenStudy (kellyspeakslouder):
ohh okay :) thank you! Do I put everything you told me from your 3rd message to your 5th message?
satellite73 (satellite73):
yes, i think so
satellite73 (satellite73):
the point is that \[\large f(g(x))=f(\frac{4}{x})=\frac{4}{\frac{4}{x}}=4\times \frac{x}{4}=x\] and since \(f(g(x))=x\) that means they are inverses
or rather "the function is its own inverse"