미분계수 정의 활용

[출제의도] 미분계수 이해하기 다항함수 $f \left( x \right)$는 $x = 1$에서 미분가능하므로 $6$$\displaystyle = \lim\limits _{h \rightarrow 0} \frac{f \left( 1 + h \right) - f \left( 1 - h \right)}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0} \left\{ \frac{f \left( 1 + h \right) - f \left( 1 \right)}{h} + \frac{f \left( 1 - h \right) - f \left( 1 \right)}{- h} \right\}$$= 2 f' \left( 1 \right)$ 에서 $f' \left( 1 \right) = 3$ $\displaystyle \lim\limits _{x \rightarrow 1} \frac{f \left( x ^{3} \right) - f \left( 1 \right)}{x - 1}$$\displaystyle = \lim\limits _{x \rightarrow 1} \left\{ \frac{f \left( x ^{3} \right) - f \left( 1 \right)}{x ^{3} - 1} \times \left( x ^{2} + x + 1 \right) \right\}$ $= f' \left( 1 \right) \times 3$$= 9$