절댓값 함수의 연속

$\left( x - 1 \right) g \left( x \right) = \left| f \left( x \right) \right|$에 $x = 1$을 대입하면 $f \left( 1 \right) = 0$ $x = 3$을 대입하면 $2 g \left( 3 \right) = \left| f \left( 3 \right) \right|$ $g \left( 3 \right) = 0$이므로 $f \left( 3 \right) = 0$ $f \left( x \right) = \left( x - 1 \right) \left( x - 3 \right) \left( x - a \right) \left( a \right.$는 상수$\left. \right)$라 하자. $\displaystyle g \left( x \right) = \frac{\left| \left( x - 1 \right) \left( x - 3 \right) \left( x - a \right) \right|}{x - 1} \left( x \ne 1 \right)$ 함수 $g \left( x \right)$가 $x = 1$에서 연속이므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} g \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 1 -} {} \frac{\left| \left( x - 1 \right) \left( x - 3 \right) \left( x - a \right) \right|}{x - 1}$$= - 2 \left| 1 - a \right|$ $\displaystyle \lim\limits _{x \rightarrow 1 +} {} g \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 1 +} {} \frac{\left| \left( x - 1 \right) \left( x - 3 \right) \left( x - a \right) \right|}{x - 1} = 2 \left| 1 - a \right|$$- 2 \left| 1 - a \right| = 2 \left| 1 - a \right|$, $a = 1$ $f \left( x \right) = \left( x - 1 \right) ^{2} \left( x - 3 \right)$이므로 $f \left( 4 \right) = 9$