역함수의 미분

함수 $f \left( x ^{3} + x \right)$의 역함수가 $g \left( x \right)$이고 $f \left( 2 \right) = 1$이므로 $g \left( 1 \right) = k$라 하면 $f \left( k ^{3} + k \right) = 1$에서 $k ^{3} + k = 2$, $k ^{3} + k - 2 = 0$ $\left( k - 1 \right) \left( k ^{2} + k + 2 \right) = 0$ $\therefore$ $k = 1 \left( \because k ^{2} + k + 2 > 0 \right)$ 즉 $g \left( 1 \right) = 1$이고 $y = f \left( x ^{3} + x \right)$라 하면 $\displaystyle \frac{dy}{dx} = \left( 3 x ^{2} + 1 \right) f' \left( x ^{3} + x \right)$ 이므로 $\displaystyle g' \left( 1 \right) = \frac{1}{\left( 3 \times 1 ^{2} + 1 \right) f' \left( 1 ^{3} + 1 \right)} = \frac{1}{4 f' \left( 2 \right)}$ 이때 모든 실수 $x$에 대하여 $f' \left( x \right) > 0$이므로 $f' \left( 2 \right) > 0$이고 $f' \left( 2 \right) = 8 g' \left( 1 \right) - 1$에서 $\displaystyle f' \left( 2 \right) = 8 \times \frac{1}{4 f' \left( 2 \right)} - 1$, $\left\{ f' \left( 2 \right) \right\} ^{2} = 2 - f' \left( 2 \right)$ $\left\{ f' \left( 2 \right) \right\} ^{2} + f' \left( 2 \right) - 2 = 0$, $\left\{ f' \left( 2 \right) - 1 \right\} \left\{ f' \left( 2 \right) + 2 \right\} = 0$ $\therefore$ $f' \left( 2 \right) = 1 \left( \because f' \left( 2 \right) > 0 \right)$ 따라서 $\displaystyle g' \left( 1 \right) = \frac{1}{4 f' \left( 2 \right)} = \frac{1}{4}$ 이므로 $\displaystyle g \left( 1 \right) + g' \left( 1 \right) = 1 + \frac{1}{4} = \frac{5}{4}$