구간별 함수 연속·미분

$\displaystyle f _{1} ( x ) = \frac{x ^{2}}{2} + x + \frac{3}{2}$, $f _{2} ( x ) = - x ^{2} + 4 x$라 두면 $f _{1}' ( x ) = x + 1 ( x < 1 )$, $f _{2}' ( x ) = - 2 x + 4 ( x \geq 1 )$ ㄱ. $f ( 1 ) = f _{1} ( 1 ) = f _{2} ( 1 ) = 3$이고, $\displaystyle \lim\limits _{x \rightarrow 1} f ( x ) = 3$ $\displaystyle \because ) \lim\limits _{x \rightarrow 1 - 0} \left( \frac{x ^{2}}{2} + x + \frac{3}{2} \right) = \lim\limits _{x \rightarrow 1 + 0} ( - x ^{2} + 4 x ) = 3$ $\displaystyle \therefore f ( 1 ) = \lim\limits _{x \rightarrow 1} f ( x )$이므로 연속이다. ㄴ. $f' ( 1 ) = 2$이므로 미분가능이다. $\because ) f _{1}' ( 1 ) = f _{2}' ( 1 ) = 2$ ㄷ. $f' ( 1 ) = f _{1}' ( 1 ) = f _{2}' ( 1 ) = 2$ 이고 $\displaystyle \lim\limits _{x \rightarrow 1} f' ( x ) = 2$ $\displaystyle \because ) \lim\limits _{x \rightarrow 1 - 0} \left( x + 1 \right) = \lim\limits _{x \rightarrow 1 + 0} ( - 2 x + 4 ) = 2$ $\displaystyle \therefore f' ( 1 ) = \lim\limits _{x \rightarrow 1} f' ( x )$이므로 연속이다.