Giải bài 27 trang 24 - SGK Giải tích lớp 12 nâng cao

Ta có: $f\left( -3 \right)=3;f\left( 1 \right)=1$

Vậy $\underset{x\in \left[ -3;1 \right]}{\mathop{Max}}\,f\left( x \right)=3;\,\underset{x\in \left[ -3;1 \right]}{\mathop{Min}}\,f\left( x \right)=1$

b)

TXĐ: $D=\left[ -2;2 \right]$

$\begin{aligned} & f'\left( x \right)=1-\dfrac{2x}{2\sqrt{4-{{x}^{2}}}}=\dfrac{\sqrt{4-{{x}^{2}}}-x}{\sqrt{4-{{x}^{2}}}} \\ & f'\left( x \right)=0\Rightarrow \sqrt{4-{{x}^{2}}}-x=0 \\ & \Leftrightarrow \left\{ \begin{aligned} & x\in \left( 0;2 \right) \\ & 4-{{x}^{2}}={{x}^{2}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x\in \left( 0;2 \right) \\ & x=\sqrt{2} \\ \end{aligned} \right. \\ \end{aligned}$

Ta có:

Vậy $\underset{x\in \left[ -2;2 \right]}{\mathop{Max}}\,f\left( x \right)=2\sqrt{2};\underset{x\in \left[ -2;2 \right]}{\mathop{Min}}\,f\left( x \right)=-2$

c)

TXĐ: $D=\mathbb{R}$

Ta có: $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{2}}x+2={{\sin }^{4}}x+1-{{\sin }^{2}}x+2={{\sin }^{4}}x-{{\sin }^{2}}x+3$

Đặt ${{\sin }^{2}}x=t\,\,\left( 0\le t\le 1 \right)$

Ta có: $g\left( t \right)={{t}^{2}}-t+3$

$\begin{aligned} & g'\left( t \right)=2t-1 \\ & g'\left( t \right)=0\Leftrightarrow t=\dfrac{1}{2} \\ \end{aligned}$

d)

$\begin{aligned} & f'\left( x \right)=1-2\cos 2x \\ & f'\left( x \right)=0\Leftrightarrow \cos 2x=\dfrac{1}{2} \\ & \Leftrightarrow \left[ \begin{aligned} & 2x=\dfrac{\pi }{3}+k2\pi \\ & 2x=-\dfrac{\pi }{3}+k2\pi \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=\dfrac{\pi }{6}+k\pi \\ & x=-\dfrac{\pi }{6}+k\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

Vì $x\in D\Rightarrow f'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned} & x=-\dfrac{\pi }{6} \\ & x=\dfrac{\pi }{6} \\ & x=\dfrac{5\pi }{6} \\ \end{aligned} \right.$

Ta có:

$\begin{aligned} & f\left( \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}-\dfrac{\sqrt{3}}{2};f\left( -\dfrac{\pi }{6} \right)=-\dfrac{\pi }{6}+\dfrac{\sqrt{3}}{2};f\left( \dfrac{5\pi }{6} \right)=\dfrac{5\pi }{6}+\dfrac{\sqrt{3}}{2} \\ & f\left( \dfrac{-\pi }{2} \right)=-\dfrac{\pi }{2};f\left( \pi \right)=\pi \\ \end{aligned}$

Vậy $\underset{x\in D}{\mathop{Max}}\,f\left( x \right)=\dfrac{5\pi }{6}+\dfrac{\sqrt{3}}{2};\underset{x\in D}{\mathop{Min}}\,f\left( x \right)=-\dfrac{\pi }{2}$