3.6-bo'lim : Funktsiyalarni birlashtirish
5. \(R\left( t \right) = \sqrt t - 2\) va \(A\left( t \right) = {\left( {t + 2} \right)^2}\) berilgan holda, \(t \ge 0\) quyidagi har birini hisoblang.
- \(\left( {R \circ A} \right)\left( t \right)\)
- \(\left( {A \circ R} \right)\left( t \right)\)
a \(\left( {R \circ A} \right)\left( t \right)\) Yechim
Esda tutingki, "\( \circ \)" ko'paytirishni emas, balki kompozitsiyani bildiradi!
\[\left( {R \circ A} \right)\left( t \right) = R\left[ {A\left( t \right)} \right] = R\left[ {{{\left( {t + 2} \right)}^2}} \right] = \sqrt {{{\left( {t + 2} \right)}^2}} - 2 = \left( {t + 2} \right) - 2 = \require{bbox} \bbox[2pt,border:1px solid black]{t}\]b \(\left( {A \circ R} \right)\left( t \right)\) Yechim
Esda tutingki, "\( \circ \)" ko'paytirishni emas, balki kompozitsiyani bildiradi!
\[\left( {A \circ R} \right)\left( t \right) = A\left[ {R\left( t \right)} \right] = A\left[ {\sqrt t - 2} \right] = {\left( {\sqrt t - 2 + 2} \right)^2} = {\left( {\sqrt t } \right)^2} = \require{bbox} \bbox[2pt,border:1px solid black]{t}\]