3.7-bo'lim : Teskari funksiyalar
6. \(\displaystyle h\left( x \right) = \frac{{1 + 2x}}{{7 + x}}\) berilgan \({h^{ - 1}}\left( x \right)\) ni toping.
Yechim
Birinchi qadam uchun funksiyani \(y\) bilan almashtiramiz.
\[y = \frac{{1 + 2x}}{{7 + x}}\]2-qadam
Keyin barcha \(x\)-larni \(y\)-ga va barcha original \(y\)-larni \(x\)-ga almashtiring.
\[x = \frac{{1 + 2y}}{{7 + y}}\]3-qadam
\(y\) uchun 2-bosqichdagi tenglamani yeching.
\[\begin{align*}x & = \frac{{1 + 2y}}{{7 + y}}\\ x\left( {7 + y} \right) & = 1 + 2y\\ 7x + xy & = 1 + 2y\\ 7x - 1 & = 2y - xy\\ 7x - 1 & = \left( {2 - x} \right)y\\ \frac{{7x - 1}}{{2 - x}} & = y\end{align*}\]4-qadam
\(y\) o'rniga \({h^{ - 1}}\left( x \right)\).
\[{h^{ - 1}}\left( x \right) = \frac{{7x - 1}}{{2 - x}}\]5-qadam
Nihoyat, \(\left( {h \circ {h^{ - 1}}} \right)\left( x \right)\) va \(\left( {{h^{ - 1}} \circ h} \right)\left( x \right)\) dan birini yoki ikkalasini hisoblash orqali tezkor tekshirib ko'ring va ularning har biri \(x\) ekanligini tekshiring. Umuman olganda, biz odatda ulardan birini tekshiramiz va biz buni shu erda qilamiz.
\[\begin{align*}\left( {h \circ {h^{ - 1}}} \right)\left( x \right) = h\left[ {{h^{ - 1}}\left( x \right)} \right] & = h\left[ {\frac{{7x - 1}}{{2 - x}}} \right]\\ & = \frac{{1 + 2\left( {\frac{{7x - 1}}{{2 - x}}} \right)}}{{7 + \frac{{7x - 1}}{{2 - x}}}}\frac{{2 - x}}{{2 - x}} = \frac{{2 - x + 2\left( {7x - 1} \right)}}{{7\left( {2 - x} \right) + 7x - 1}} = \frac{{13x}}{{13}} = x\end{align*}\]Tekshirish amalga oshiriladi, shuning uchun biz ishni to'g'ri bajarganimizni va teskarisini bilamiz.