Keyin barcha \(x\)-larni \(y\)-ga va barcha original \(y\)-larni \(x\)-ga almashtiring.

\[x = \sqrt[5]{{2y + 11}}\]

\(y\) uchun 2-bosqichdagi tenglamani yeching.

\[\begin{align*}{x^5} & = 2y + 11\\ {x^5} - 11 & = 2y\\\frac{{{x^5} - 11}}{2} & = y\end{align*}\]

\(y\) o'rniga \({A^{ - 1}}\left( x \right)\).

\[{A^{ - 1}}\left( x \right) = \frac{{{x^5} - 11}}{2}\]

Nihoyat, \(\left( {A \circ {A^{ - 1}}} \right)\left( x \right)\) va \(\left( {{A^{ - 1}} \circ A} \right)\left( x \right)\) 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.

\[\begin{align*}\left( {A \circ {A^{ - 1}}} \right)\left( x \right) = A\left[ {{A^{ - 1}}\left( x \right)} \right] & = A\left[ {\frac{{{x^5} - 11}}{2}} \right]\\ & = \sqrt[5]{{2\left( {\frac{{{x^5} - 11}}{2}} \right) + 11}} = \sqrt[5]{{{x^5} - 11 + 11}} = \sqrt[5]{{{x^5}}} = x\end{align*}\]

Tekshirish amalga oshiriladi, shuning uchun biz ishni to'g'ri bajarganimizni va teskarisini bilamiz.