1.2-bo'lim : Teskari funksiyalar
3. \(R\left( x \right) = {x^3} + 6\) uchun teskarisini toping. Ushbu bo'limda muhokama qilinganidek, kompozitsiyadan birini yoki ikkalasini hisoblash orqali teskarisini tekshiring.
Maslahat: Ushbu bo'limda tasvirlangan jarayonni eslang. \(R\left( x \right)\)-ni almashtiring, \(x\) va \(y\)-ni almashtiring, \(y\)-ni hal qiling va nihoyat \(y\)-ni \({R^{ - 1}}\left( x \right)\) bilan almashtiring.
Yechim
\[y = {x^3} + 6\]
2-qadam
\[x = {y^3} + 6\]
3-qadam
\[\begin{align*}x - 6 & = {y^3}\\ y &= \sqrt[3]{{x - 6}}\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{R^{ - 1}}\left( x \right) = \sqrt[3]{{x - 6}}}}\end{align*}\]
Va nihoyat, ishimizni tekshirish uchun \(\left( {R \circ {R^{ - 1}}} \right)\left( x \right)\) yoki \(\left( {{R^{ - 1}} \circ R} \right)\left( x \right)\) hisoblang.
4-qadam
Ikkala kompozitsiya ham bajarilishi mumkin, keling, bu holda \(\left( {{R^{ - 1}} \circ R} \right)\left( x \right)\) ni bajaraylik.
\[\begin{align*}\left( {{R^{ - 1}} \circ R} \right)\left( x \right) & = {R^{ - 1}}\left[ {R\left( x \right)} \right]\\ & = \sqrt[3]{{\left( {{x^3} + 6} \right) - 6}}\\ & = \sqrt[3]{{{x^3}}}\\ & = x\end{align*}\]Shunday qilib, biz \(x\)-ni kompozitsiyadan chiqardik va shuning uchun biz o'z ishimizni to'g'ri bajarganimizni bilamiz.