절댓값 함수 미분가능

함수 $g \left( x \right)$가 $x = 0$에서 연속이므로 $\displaystyle \lim\limits _{x \rightarrow 0 -} g \left( x \right) = - f \left( 0 \right)$, $\displaystyle \lim\limits _{x \rightarrow 0 +} g \left( x \right) = \left| f \left( 0 \right) \right| - 8$ $- f \left( 0 \right) = \left| f \left( 0 \right) \right| - 8$ $f \left( 0 \right) = 4$ $\cdots$$\cdots$ ㉠ $g \left( 0 \right) = \left| f \left( 0 \right) \right| - 8 = - 4$ 함수 $g \left( x \right)$가 $x = 0$에서 미분가능하므로 $\displaystyle \lim\limits _{h \rightarrow 0 -} \frac{g \left( 0 + h \right) - g \left( 0 \right)}{h}$$\displaystyle = \lim\limits _{h \rightarrow 0 -} \frac{- f \left( 0 + h \right) + f \left( 0 \right)}{h}$$= - f' \left( 0 \right)$ $\displaystyle \lim\limits _{h \rightarrow 0 +} \frac{g \left( 0 + h \right) - g \left( 0 \right)}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} \frac{\left| f \left( 0 + h \right) \right| + 2 h ^{2} - 8 + 4}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} \frac{\left| f \left( 0 + h \right) \right| - f \left( 0 \right)}{h} + \lim\limits _{h \rightarrow 0 +} \frac{2 h ^{2}}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} {} \frac{f \left( 0 + h \right) - f \left( 0 \right)}{h} + 0 = f' \left( 0 \right)$ $- f' \left( 0 \right) = f' \left( 0 \right)$ $f' \left( 0 \right) = 0$ $\cdots$$\cdots$ ㉡ 함수 $g \left( x \right)$가 $x = 2$에서 미분가능하므로 $\displaystyle \lim\limits _{h \rightarrow 0 -} \frac{g \left( 2 + h \right) - g \left( 2 \right)}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 -} \frac{\left| f \left( 2 + h \right) \right| + 2 \left( 2 + h \right) ^{2} - 8 - \left| f \left( 2 \right) \right|}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 -} \frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h} + \lim\limits _{h \rightarrow 0 -} \frac{2 h ^{2} + 8 h}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 -} \frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h} + 8$ $\displaystyle \lim\limits _{h \rightarrow 0 +} \frac{g \left( 2 + h \right) - g \left( 2 \right) - g \left( 2 \right)}{h}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} {\frac{\left| f \left( 2 + h \right) \right| - 2 \left( 2 + h \right) ^{2} + 8 - \left| f \left( 2 \right) \right|}{h}}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} {\frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h}} - \lim\limits _{h \rightarrow 0 +} {\frac{2 h ^{2} + 8 h}{h}}$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} {\frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h}} - 8$ $\displaystyle \lim\limits _{h \rightarrow 0 -} {\frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h}} + 8$ $\displaystyle = \lim\limits _{h \rightarrow 0 +} {\frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h}} - 8$ $\cdots$$\cdots$ ㉢ 즉 함수 $\left| f \left( x \right) \right|$가 $x = 2$에서 미분가능하지 않다. $f \left( 2 \right) = 0$ $\cdots$$\cdots$ ㉣ $\displaystyle \lim\limits _{h \rightarrow 0 +} \frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h} = \left| f' \left( 2 \right) \right|$와 $\displaystyle \lim\limits _{h \rightarrow 0 -} \frac{\left| f \left( 2 + h \right) \right| - \left| f \left( 2 \right) \right|}{h} = - \left| f' \left( 2 \right) \right|$를 ㉢에 대입하면 $\left| f' \left( 2 \right) \right| = 8$ $\cdots$$\cdots$ ㉤ ㉠, ㉡, ㉣, ㉤을 모두 만족시키는 함수 $f \left( x \right)$는 $f \left( x \right) = 3 x ^{3} - 7 x ^{2} + 4$ 또는 $f \left( x \right) = - x ^{3} + x ^{2} + 4$ (ⅰ) $f \left( x \right) = 3 x ^{3} - 7 x ^{2} + 4$인 경우 $3 x ^{3} - 7 x ^{2} + 4 = \left( 3 x + 2 \right) \left( x - 1 \right) \left( x - 2 \right)$이므로 함수 $g \left( x \right)$가 $x = 1$에서 미분가능하지 않다. (ⅱ) $f \left( x \right) = - x ^{3} + x ^{2} + 4$인 경우 $- x ^{3} + x ^{2} + 4 = - \left( x - 2 \right) \left( x ^{2} + x + 2 \right)$이므로 함수 $g \left( x \right)$는 실수 전체의 집합에서 미분가능하다. (ⅰ), (ⅱ)에서 $f \left( x \right) = - x ^{3} + x ^{2} + 4$이다. 따라서 $f \left( - 5 \right) = 154$