1.2-bo'lim : Teskari funksiyalar
6. \(f\left( x \right) = \sqrt[7]{{5x + 8}}\) uchun teskarisini toping. Ushbu bo'limda muhokama qilinganidek, kompozitsiyadan birini yoki ikkalasini hisoblash orqali teskarisini tekshiring.
Maslahat: Ushbu bo'limda tasvirlangan jarayonni eslang. \(f\left( x \right)\)-ni almashtiring, \(x\) va \(y\)-ni almashtiring, \(y\)-ni hal qiling va nihoyat \(y\)-ni \({f^{ - 1}}\left( x \right)\) bilan almashtiring.
Yechim
\[y = \sqrt[7]{{5x + 8}}\]
2-qadam
\[x = \sqrt[7]{{5y + 8}}\]
3-qadam
\[\begin{align*}x & = \sqrt[7]{{5y + 8}}\\ {x^7} & = 5y + 8\\ & {x^7} - 8 = 5y\\ y & = \frac{1}{5}\left( {{x^7} - 8} \right)\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{f^{ - 1}}\left( x \right) = \frac{1}{5}\left( {{x^7} - 8} \right)}}\end{align*}\]
Va nihoyat, ishimizni tekshirish uchun \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) yoki \(\left( {{f^{ - 1}} \circ f} \right)\left( x \right)\) hisoblang.
4-qadam
Ikkala kompozitsiya ham bajarilishi mumkin, keling, bu holda \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) ni bajaraylik.
\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) & = f\left[ {{f^{ - 1}}\left( x \right)} \right]\\ & = \sqrt[7]{{5\left[ {\frac{1}{5}\left( {{x^7} - 8} \right)} \right] + 8}}\\ & = \sqrt[7]{{\left[ {{x^7} - 8} \right] + 8}}\\ & = \sqrt[7]{{{x^7}}}\\ & = x\end{align*}\]Shunday qilib, biz \(x\)-ni kompozitsiyadan chiqardik va shuning uchun biz o'z ishimizni to'g'ri bajarganimizni bilamiz.