Fuincseanan co-cheangailte adhartach
Ma tha thu a' faireachdainn misneachail, feuch an t-eisimpleir seo a tha nas dorra. Na gabh dragh ma tha e a' coimhead doirbh oir bidh thu a' leantainn an aon phròiseas 's a bha thu roimhe. Cuir am fuincsean an àite \(x\) agus an uair sin sìmplich e.
Question
\(f(x) = \frac{{x - 1}}{{x + 1}}\)
Obraich a-mach \(h(x) = f(f(x))\)
\(f(f(x))=\frac{f(x)-1}{f(x)+1}\)
\( = \frac{{\frac{{x - 1}}{{x + 1}} - 1}}{{\frac{{x - 1}}{{x + 1}} + 1}}\)
Feumaidh tu a-nis an t-àireamhaiche agus an seòrsaiche a shìmpleachadh air leth.
Àireamhaiche
\(\frac{{x - 1}}{{x + 1}} - 1\)
\(= \frac{{x - 1}}{{x + 1}} - \frac{{x + 1}}{{x + 1}}\)
\(= \frac{{x - 1 - (x + 1)}}{{x + 1}}\)
\(= \frac{{ - 2}}{{x + 1}}\)
Seòrsaiche
\(\frac{{x - 1}}{{x + 1}} + 1\)
\(=\frac{{x - 1}}{{x + 1}} + \frac{{x + 1}}{{x + 1}}\)
\(= \frac{{x - 1 + x + 1}}{{x + 1}}\)
\(= \frac{{2x}}{{x + 1}}\)
Tha seo againn a-nis:
\(f(f(x))=\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}\)
\(= \frac{{ - 2}}{{x + 1}} \div \frac{{2x}}{{x + 1}}\)
\(= \frac{{ - 2}}{{x + 1}} \times \frac{{x + 1}}{{2x}}\)
\(= \frac{{ - 2(x + 1)}}{{2x(x + 1)}}\)
\(= \frac{{ - 2}}{{2x}}\)
\(= \frac{{ - 1}}{x}\)
Mar sin:
\(h(x) = \frac{{ - 1}}{x}\)
