적분함수 최댓값 보기

ㄱ. $f' \left( x \right) = x ^{2} - 4 tx + 3 t ^{2}$$= \left( x - t \right) \left( x - 3 t \right)$ (참) ㄴ. $f \left( x \right)$$\displaystyle = \int _{3 t} ^{x} {\left( s ^{2} - 4 ts + 3 t ^{2} \right) ds}$ $\displaystyle = \left[ \frac{1}{3} s ^{3} - 2 ts ^{2} + 3 t ^{2} s \right] _{3 t} ^{x}$$\displaystyle = \frac{1}{3} x ^{3} - 2 tx ^{2} + 3 t ^{2} x$ $\displaystyle = \frac{1}{3} x \left( x - 3 t \right) ^{2}$ $f' \left( t \right) = 0$, $\displaystyle f ( t ) = \frac{4}{3} t ^{3}$이므로 $\displaystyle f \left( x \right) - \frac{4}{3} t ^{3} = \frac{1}{3} x \left( x ^{2} - 6 tx + 9 t ^{2} \right) - \frac{4}{3} t ^{3}$$\displaystyle = \frac{1}{3} \left( x - t \right) ^{2} \left( x - 4 t \right)$ $\displaystyle f \left( t \right) = f \left( 4 t \right) = \frac{4}{3} t ^{3}$ ⅰ) $t > 2$일 때

$\displaystyle \frac{4}{3} t ^{3}$ $y = f \left( x \right)$

함수 $f \left( x \right)$는 $x = 2$에서 최댓값을 가지므로 $\displaystyle g ( t ) = f \left( 2 \right) = \frac{2}{3} ( 3 t - 2 ) ^{2}$ ⅱ) $t \leq 2 < 4 t$ 즉, $\displaystyle \frac{1}{2} < t \leq 2$일 때

$\displaystyle \frac{4}{3} t ^{3}$ $y = f \left( x \right)$

함수 $f ( x )$는 $x = t$에서 최댓값을 가지므로 $\displaystyle g ( t ) = f \left( t \right) = \frac{4}{3} t ^{3}$ ⅲ) $4 t \leq 2$ 즉, $\displaystyle 0 < t \leq \frac{1}{2}$일 때

$\displaystyle \frac{4}{3} t ^{3}$ $y = f \left( x \right)$

함수 $f ( x )$는 $x = 2$에서 최댓값을 가지므로 $\displaystyle g ( t ) = f ( 2 ) = \frac{2}{3} ( 3 t - 2 ) ^{2}$ ⅰ), ⅱ), ⅲ)에 의하여 $t > 2$일 때, $\displaystyle g ( t ) = \frac{2}{3} ( 3 t - 2 ) ^{2}$ (참) ㄷ. 함수

의 미분가능성을 조사하면 ⅰ) $\displaystyle t = \frac{1}{2}$일 때, $\displaystyle \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} -} {\frac{g ( t ) - g \left( \frac{1}{2} \right)}{t - \frac{1}{2}}}$$\displaystyle = \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} -} {\frac{\frac{2}{3} ( 3 t - 2 ) ^{2} - \frac{1}{6}}{t - \frac{1}{2}}}$ $\displaystyle = \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} -} {\frac{( 2 t - 1 ) ( 6 t - 5 )}{2 t - 1}}$$= - 2$ $\displaystyle \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} +} {\frac{g ( t ) - g \left( \frac{1}{2} \right)}{t - \frac{1}{2}}}$$\displaystyle = \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} +} {\frac{\frac{4}{3} t ^{3} - \frac{1}{6}}{t - \frac{1}{2}}}$ $\displaystyle = \lim\limits _{\mathit{t} \rightarrow \frac{1}{2} +} {\frac{\left( 2 t - 1 ^{} \right) \left( 4 t ^{2} + 2 t + 1 \right)}{3 ( 2 t - 1 )}}$$= 1$ 따라서 $\displaystyle t = \frac{1}{2}$에서 미분가능하지 않다. ⅱ) $t = 2$일 때, $\displaystyle \lim\limits _{\mathit{t} \rightarrow 2 -} {\frac{g ( t ) - g ( 2 )}{t - 2}}$$\displaystyle = \lim\limits _{\mathit{t} \rightarrow 2 -} {\frac{\frac{4}{3} t ^{3} - \frac{32}{3}}{t - 2}}$ $\displaystyle = \lim\limits _{\mathit{t} \rightarrow 2 -} {\frac{\frac{4}{3} ( t ^{3} - 8 )}{t - 2}}$$\displaystyle = \lim\limits _{\mathit{t} \rightarrow 2 -} {\frac{4 ( t ^{2} + 2 t + 4 )}{3}} = 16$ $\displaystyle \lim\limits _{\mathit{t} \rightarrow 2 +} {\frac{g ( t ) - g ( 2 )}{t - 2}}$$\displaystyle = \lim\limits _{\mathit{t} \rightarrow 2 +} {\frac{\frac{2}{3} ( 3 t - 2 ) ^{2} - \frac{32}{3}}{t - 2}}$ $\displaystyle = \lim\limits _{\mathit{t} \rightarrow 2 +} {\frac{2 ( t - 2 ) ( 3 t + 2 )}{t - 2}} = 16$ 따라서 $t = 2$에서 미분가능하다. ⅰ), ⅱ)에 의하여 $t > 0$에서 함수 $g \left( t \right)$는 $\displaystyle t = \frac{1}{2}$에서만 미분가능하지 않다. (참) 따라서 옳은 것은 ㄱ, ㄴ, ㄷ