곱함수 미분가능

[출제의도] 미분계수 추론하기 $h' \left( 4 \right)$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{f \left( x \right) g \left( x \right) - f \left( 4 \right) g \left( 4 \right)}{x - 4}}$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{f \left( x \right) \times \frac{1}{x - 4} - 2 f \left( 4 \right)}{x - 4}}$ $\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{f \left( x \right) - 2 \left( x - 4 \right) f \left( 4 \right)}{\left( x - 4 \right) ^{2}}}$ $\cdots \text{㉠}$ 에서 $\displaystyle \lim\limits _{x \rightarrow 4} {\left( x - 4 \right) ^{2} = 0}$이므로 $\displaystyle \lim\limits _{x \rightarrow 4} {\left\{ f \left( x \right) - 2 \left( x - 4 \right) f \left( 4 \right) \right\} = f \left( 4 \right) = 0}$ $f \left( x \right) = \left( x - 4 \right) \left( x ^{2} + ax + b \right)$라 두고 $\text{㉠}$에 대입하면 $h' \left( 4 \right)$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{\left( x - 4 \right) \left( x ^{2} + ax + b \right)}{\left( x - 4 \right) ^{2}}}$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{x ^{2} + ax + b}{x - 4}}$ $\cdots \text{㉡}$ 이고 $\displaystyle \lim\limits _{x \rightarrow 4} {\left( x - 4 \right) = 0}$이므로 $\displaystyle \lim\limits _{x \rightarrow 4} {\left( x ^{2} + ax + b \right) = 16 + 4 a + b = 0}$ $b = \mathit{-} 4 a - 16$이고 이를 $\text{㉡}$에 대입하면 $h' \left( 4 \right)$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{x ^{2} + ax - 4 a - 16}{x - 4}}$$\displaystyle = \lim\limits _{x \rightarrow 4} {\frac{\left( x - 4 \right) \left( x + a + 4 \right)}{x - 4}}$ $= a + 8 = 6$ 에서 $a = \mathit{-} 2$, $b = \mathit{-} 8$이다. $\therefore$ $f \left( x \right)$$= \left( x - 4 \right) ^{2} \left( x + 2 \right)$ 따라서 $f \left( 0 \right) = 16 \times 2 = 32$