절댓값 함수 미분

함수 $\displaystyle f \left( x \right) = \begin{cases} - ax + 2 a & & \left( x < 2 \right) \\ ax - 2 a & & \left( x \geq 2 \right) \end{cases}$ 에 대하여 (ⅰ) $k < 2$일 때 $\displaystyle \lim\limits _{x \rightarrow k} \frac{f \left( x \right) - f \left( k \right)}{x - k}$$\displaystyle = \lim\limits _{x \rightarrow k} \frac{- ax + 2 a - \left( - ak + 2 a \right)}{x - k}$ $\displaystyle = \lim\limits _{x \rightarrow k} \frac{- a \left( x - k \right)}{x - k} = - a$ 이므로 $\displaystyle \lim\limits _{x \rightarrow k +} \frac{f \left( x \right) - f \left( k \right)}{x - k} - \lim\limits _{x \rightarrow k -} \frac{f \left( x \right) - f \left( k \right)}{x - k}$ $= - a - \left( - a \right) = 0 \neq 6$ (ⅱ) $k = 2$일 때 $\displaystyle \lim\limits _{x \rightarrow 2 +} \frac{f \left( x \right) - f \left( 2 \right)}{x - 2} - \lim\limits _{x \rightarrow 2 -} \frac{f \left( x \right) - f \left( 2 \right)}{x - 2}$ $\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{ax - 2 a}{x - 2} - \lim\limits _{x \rightarrow 2 -} \frac{- ax + 2 a}{x - 2}$ $= a - \left( - a \right) = 2 a$ $2 a = 6$에서 $a = 3$ (ⅲ) $k > 2$일 때 $\displaystyle \lim\limits _{x \rightarrow k} \frac{f \left( x \right) - f \left( k \right)}{x - k}$$\displaystyle = \lim\limits _{x \rightarrow k} \frac{ax - 2 a - \left( ak - 2 a \right)}{x - k}$ $\displaystyle = \lim\limits _{x \rightarrow k} \frac{a \left( x - k \right)}{x - k} = a$ 이므로 $\displaystyle \lim\limits _{x \rightarrow k +} \frac{f \left( x \right) - f \left( k \right)}{x - k} - \lim\limits _{x \rightarrow k -} \frac{f \left( x \right) - f \left( k \right)}{x - k}$ $= a - a = 0 \neq 6$ (ⅰ), (ⅱ), (ⅲ)에 의하여 $a = 3$, $k = 2$이므로 $f \left( a + k \right) = f \left( 3 + 2 \right) = f \left( 5 \right) = 9$