극한 조건과 이차함수

[출제의도] 미분계수 이해하기 $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x \right) - g \left( x \right)}{x - 3}$$= 1$이고 $\displaystyle \lim\limits _{x \rightarrow 3} {} \left( x - 3 \right)$$= 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 3} {} \left\{ f \left( x \right) - g \left( x \right) \right\}$$= 0$ $f \left( x \right)$, $g \left( x \right)$가 모두 다항함수이므로 $f \left( 3 \right)$$= g \left( 3 \right)$이고 $f \left( 3 \right)$$= 2$이므로 $g \left( 3 \right)$$= 2$$= \mathit{-} 2$ $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x \right) - g \left( x \right)}{x - 3}$$\displaystyle = \lim\limits _{x \rightarrow 3} {} \frac{\left\{ f \left( x \right) - f \left( 3 \right) \right\} - \left\{ g \left( x \right) - g \left( 3 \right) \right\}}{x - 3}$ $= f' \left( 3 \right) - g' \left( 3 \right)$$= 1$ $f' \left( 3 \right)$$= 1$이므로 $g' \left( 3 \right)$$= 0$ $g \left( x \right)$$= x ^{2} + ax + b$($a , b$는 상수)라 하면 $g' \left( x \right)$$= 2 x + a$ $g \left( 3 \right)$$= 9 + 3 a + b$$= 2$, $g' \left( 3 \right)$$= 6 + a$$= 0$ 에서 $a$$= \mathit{-} 6$, $b = 11$ 따라서 $g \left( 1 \right)$$= 1 - 6 + 11$$= 6$