함수방정식과 도함수의 성질

$\displaystyle \lim\limits _{x} \rightarrow 0 \frac{g ( x ) - 1}{x} = 0$에서 $g ( 0 ) = 1$, $g' ( 0 ) = 0$ $f ( 1 + 0 ) = f ( 1 ) g ( 0 ) + f ( 0 ) g ( 1 )$ 에서 $f ( 0 ) g ( 1 ) = 0$ $g ( 1 + 0 ) = g ( 1 ) g ( 0 ) + f ( 1 ) f ( 0 )$ 에서 $f ( 1 ) f ( 0 ) = 0$ $f ( 1 ) = 1$이므로 $f ( 0 ) = 0$ ㄱ. $\displaystyle f' ( x ) = \lim\limits _{y \rightarrow 0} {} \frac{f ( x + y ) - f ( x )}{y}$ $\displaystyle = \lim\limits _{y \rightarrow 0} {} \frac{f ( x ) g ( y ) + f ( y ) g ( x ) - f ( x )}{y}$ $\displaystyle = \lim\limits _{y \rightarrow 0} {} f ( x ) \frac{g ( y ) - 1}{y} + \lim\limits _{y \rightarrow 0} {} \frac{f ( y )}{y} g ( x )$ $= f' ( 0 ) g ( x )$ -(a) 추가로 $\displaystyle g' ( x ) = \lim\limits _{y \rightarrow 0} {} \frac{g ( x + y ) - g ( x )}{y}$ $\displaystyle = \lim\limits _{y \rightarrow 0} {} \frac{g ( x ) g ( y ) + f ( y ) f ( x ) - g ( x )}{y}$ $\displaystyle = \lim\limits _{y \rightarrow 0} {} g ( x ) \frac{g ( y ) - 1}{y} + \lim\limits _{y \rightarrow 0} {} \frac{f ( y )}{y} f ( x )$ $= f' ( 0 ) f ( x )$ -(b) ㄷ. (a),(b)에서 $\displaystyle f' ( 0 ) = \frac{g' ( x )}{f ( x )} = \frac{f' ( x )}{g ( x )}$ 정리하면 $2 g ( x ) g' ( x ) - 2 f ( x ) f' ( x ) = 0$ 적분하면 $\{ g ( x ) \} ^{2} - \{ f ( x ) \} ^{2} = 1$ $( \because g ( 0 ) = 1 , f ( 0 ) = 0 )$ ㄴ. ㄷ에서 $\{ g ( x ) \} ^{2} = 1 + \{ f ( x ) \} ^{2} \geq 1$이므로 $x = 0$를 포함하는 열린구간에서 $0 < g ( x ) \leq 1$인 점이 없다. 따라서 이 열린구간에서 $g ( x ) \geq 1 = g ( 0 )$이므로 $x = 0$에서 극솟값 1 을 갖는다.