3.7-bo'lim : Teskari funksiyalar
5. Berilgan \(\displaystyle f\left( x \right) = \frac{{4x}}{{5 - x}}\) \({f^{ - 1}}\left( x \right)\) ni toping.
Yechim
Birinchi qadam uchun funksiyani \(y\) bilan almashtiramiz.
\[y = \frac{{4x}}{{5 - x}}\]2-qadam
Keyin barcha \(x\)-larni \(y\)-ga va barcha original \(y\)-larni \(x\)-ga almashtiring.
\[x = \frac{{4y}}{{5 - y}}\]3-qadam
\(y\) uchun 2-bosqichdagi tenglamani yeching.
\[\begin{align*}x & = \frac{{4y}}{{5 - y}}\\ x\left( {5 - y} \right) & = 4y\\ 5x - xy & = 4y\\ 5x & = 4y + xy\\ 5x & = \left( {4 + x} \right)y\\ \frac{{5x}}{{4 + x}} & = y\end{align*}\]4-qadam
\(y\) o'rniga \({f^{ - 1}}\left( x \right)\).
\[{f^{ - 1}}\left( x \right) = \frac{{5x}}{{4 + x}}\]5-qadam
Nihoyat, \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) va \(\left( {{f^{ - 1}} \circ f} \right)\left( x \right)\) 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 buni bu erda qilamiz.
\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f\left[ {{f^{ - 1}}\left( x \right)} \right] & = f\left[ {\frac{{5x}}{{4 + x}}} \right]\\ & = \frac{{4\left( {\frac{{5x}}{{4 + x}}} \right)}}{{5 - \frac{{5x}}{{4 + x}}}}\frac{{4 + x}}{{4 + x}} = \frac{{20x}}{{5\left( {4 + x} \right) - 5x}} = \frac{{20x}}{{20}} = x\end{align*}\]Tekshirish amalga oshiriladi, shuning uchun biz ishni to'g'ri bajarganimizni va teskarisini bilamiz.