극한 조건 사차함수

$\displaystyle \lim\limits _{x \rightarrow k} \frac{\left( f \left( x \right) \right) ^{2} - \left( f \left( k \right) \right) ^{2}}{x - k}$ $\displaystyle = \lim\limits _{x \rightarrow k} \left\{ \left( f \left( x \right) + f \left( k \right) \right) \times \frac{f \left( x \right) - f \left( k \right)}{x - k} \right\}$$= 2 f \left( k \right) f' \left( k \right)$ 이므로 $\displaystyle \lim\limits _{x \rightarrow k} \frac{2 x ^{2} f \left( x \right) - \left( f \left( k \right) \right) ^{2}}{x - k}$의 값이 존재한다. $\displaystyle \lim\limits _{x \rightarrow k} \left( x - k \right) = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow k} \left\{ 2 x ^{2} f \left( x \right) - \left( f \left( k \right) \right) ^{2} \right\} = 0$ $2 k ^{2} f \left( k \right) - \left( f \left( k \right) \right) ^{2} = 0$, $f \left( k \right) \left( 2 k ^{2} - f \left( k \right) \right) = 0$ 이므로 $f \left( k \right) = 0$ 또는 $f \left( k \right) = 2 k ^{2}$ 함수 $f \left( x \right)$의 최솟값이 $17$이므로 $f \left( k \right) = 2 k ^{2}$ $\displaystyle \lim\limits _{x \rightarrow k} \frac{\left( f \left( x \right) \right) ^{2} - \left( f \left( k \right) \right) ^{2}}{x - k} = 4 k ^{2} f' \left( k \right)$ $\displaystyle \lim\limits _{x \rightarrow k} \frac{2 x ^{2} f \left( x \right) - \left( f \left( k \right) \right) ^{2}}{x - k}$ $\displaystyle = \lim\limits _{x \rightarrow k} \frac{2 x ^{2} f \left( x \right) - 2 k ^{2} f \left( k \right)}{x - k}$ $\displaystyle = \lim\limits _{x \rightarrow k} \left( 2 x ^{2} \times \frac{f \left( x \right) - f \left( k \right)}{x - k} + f \left( k \right) \times \frac{2 x ^{2} - 2 k ^{2}}{x - k} \right)$ $= 2 k ^{2} f' \left( k \right) + 4 kf \left( k \right)$ $= 2 k ^{2} f' \left( k \right) + 8 k ^{3}$ $2 k ^{2} f' \left( k \right) + 8 k ^{3} = 4 k ^{2} f' \left( k \right)$ $2 k ^{2} \left( f' \left( k \right) - 4 k \right) = 0$ $k \neq 0$이므로 $f' \left( k \right) = 4 k$ $k = t$일 때, $f \left( t \right) = 2 t ^{2}$, $f' \left( t \right) = 4 t$ $k = - t$일 때, $f \left( - t \right) = 2 t ^{2}$, $f' \left( - t \right) = - 4 t$ $f \left( t \right) = 2 t ^{2}$, $f \left( - t \right) = 2 t ^{2}$이므로 두 실수 $a$, $b$에 대하여 $f \left( x \right) - 2 t ^{2}$$= \left( x - t \right) \left( x + t \right) \left( x ^{2} + ax + b \right)$ $= \left( x ^{2} - t ^{2} \right) \left( x ^{2} + ax + b \right)$ $f' \left( x \right) = 2 x \left( x ^{2} + ax + b \right) + \left( x ^{2} - t ^{2} \right) \left( 2 x + a \right)$ $f' \left( t \right) = 4 t$, $f' \left( - t \right) = - 4 t$이므로 $f' \left( t \right) = 2 t \left( t ^{2} + at + b \right) = 4 t$ $f' \left( - t \right) = - 2 t \left( t ^{2} - at + b \right) = - 4 t$ $t > 1$이므로 $t ^{2} + at + b = 2$, $t ^{2} - at + b = 2$ 에서 $a = 0$, $b = - t ^{2} + 2$ 그러므로 $f \left( x \right)$$= \left( x ^{2} - t ^{2} \right) \left( x ^{2} - t ^{2} + 2 \right) + 2 t ^{2}$ $= \left( x ^{2} - t ^{2} + 1 \right) ^{2} + 2 t ^{2} - 1$ 함수 $f \left( x \right)$의 최솟값은 $2 t ^{2} - 1$이다. $2 t ^{2} - 1 = 17$, $t = 3$ 따라서 $f \left( x \right) = \left( x ^{2} - 8 \right) ^{2} + 17$이므로 $f \left( 4 \right) = 81$