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