第7章 7.29

(1)

$y = \sin3x + 3\sin x$

$y' = 3\cos3x + 3\cos x$

$y'=0$となるのは $3\cos3x + 3\cos x = 0$ $\cos3x + \cos x = 0$

$\cos(\alpha+\beta) + \cos(\alpha-\beta) = 2\cos\alpha\cdot\cos\beta$から

$\cos3x + \cos x = 2\cos2x\cdot\cos x = 0$ $\cos2x = 0$、 $\cos x = 0$ $x = \dfrac{\pi}{4}$、 $\dfrac{1}{2}\pi$

範囲は$\left(-\dfrac{1}{6}\pi \le x \le \dfrac{\pi}{3}\right)$なので $x = \dfrac{\pi}{4}$

$x = -\dfrac{\pi}{6}$のとき $y = \sin3\cdot\left(-\dfrac{\pi}{6}\right) + 3\sin\left(-\dfrac{\pi}{6}\right) = -1 + 3\cdot\left(-\dfrac{1}{2}\right) = -\dfrac{5}{2}$

$x = \dfrac{\pi}{4}$のとき $y = \sin3\cdot\left(\dfrac{\pi}{4}\right) + 3\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} + 3\cdot\dfrac{1}{\sqrt{2}} = 2\sqrt{2}$

$x = \dfrac{\pi}{3}$のとき $y = \sin3\cdot\left(\dfrac{\pi}{3}\right) + 3\sin\left(\dfrac{\pi}{3}\right) = 0 + 3\cdot\dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2}$

  極大

これから、$x = \dfrac{\pi}{4}$のとき最大値$2\sqrt{2}$ $x = -\dfrac{\pi}{6}$のとき最小値$-\dfrac{5}{2}$

(2)

$y = 2^{x+2} - 4^x = 4 \cdot 2^x - 4^x$

$y' = 4 \cdot 2^x \log 2 - 4^x \log 4 = 4 \cdot 2^x \log 2 - 4^x \cdot 2\log 2 = (4 \cdot 2^x - 2 \cdot 4^x)\log 2 = (2^{x+2} - 2^{2x+1})\log 2$

$y' = 0$となるのは $2^{x+2} - 2^{2x+1} = 0$ $(x+2)\log 2 - (2x+1)\log 2 = 0$

$x+2-2x-1=0$ $-x+1=0$ $\therefore x=1$

$x=1$のとき $y=4 \cdot 2-4=8-4=4$

$x=3$のとき $y=4 \cdot 2^3-4^3=4 \cdot 8-64=-32$

極大

これから$x=1$のとき最大値$4$、$x=3$のとき最小値$-32$