関数の最大値と最小値を求める

\begin{align*}
f(x) &= (1+\sin{x}) \times \cos{x}\color{red}\scriptsize\Leftarrow A \times B \\
\color{red}\scriptsize A \times B & \color{red}\scriptsize =　　　A'　　　　B　\ +\ 　　　A　　　　B'\\
f'(x) &= (1+\sin{x})'\cos{x} + (1+\sin{x})(\cos{x})'\\
\\
&= \cos{x} \cos{x} + (1+\sin{x}) \times (-\sin{x})\\
\\
&= \cos^2{x}-\sin^2{x}-\sin{x} \color{red}\scriptsize\ 　\sin,\ \cos どちらかに統一\\
& \color{red}\scriptsize　　　\Darr \sin^2{x}+\cos^2{x}=1\ より\ \cos^2{x} = 1 - \sin^2{x} \\
&= (1-\sin^2{x})-\sin^2{x}-\sin{x}\\
&\\
&= -2\sin^2{x}-\sin{x}+1\\
\\
&= -(2\sin^2{x}+\sin{x}-1)\\
\\
&= -(2\sin{x}-1)(\sin{x}+1)
\end{align*}

\begin{align*}
f'(x) &= 0\\
-(2\sin{x}-1)(\sin{x}+1) &= 0\\
\color{red}\scriptsize 2\sin{x}-1 = 0, & 　\color{red}\scriptsize\sin{x}+1=0\\
\sin{x} &= -1,\ \dfrac12\\
x &= \dfrac32\pi,\ \dfrac16\pi,\ \dfrac56\pi\\
\end{align*}

\def\arraystretch{1.8}
\begin{array}{c||c|c|c|c|c|c|c|c|c}\hline
x & 0 & \cdots & \frac{\pi}{6} & \cdots & \frac56\pi & \cdots & \frac32\pi & \cdots & 2\pi \\\hline
f'(x) & \color{red}{\times} & + & 0 & - & 0 & + & 0 & + & \color{red}\times \\\hline
f(x) & 1 & \nearrow & \colorbox{mistyrose}{$\substack{極大\\\frac{3\sqrt{3}}{4}}$} & \searrow & \colorbox{lightcyan}{$\substack{極小\\-\frac{3\sqrt{3}}{4}}$} & \nearrow & 0 & \nearrow & 1 \\\hline
\end{array}

\begin{align*}
f(x) &= (1+\cos{x}) \times \sin{x}\color{red}\scriptsize\Leftarrow A \times B \\
\color{red}\scriptsize A \times B & \color{red}\scriptsize =　　　A'　　　　B　\ +\ 　　　A　　　　B'\\
f'(x) &= (1+\cos{x})'\sin{x} + (1+\cos{x})(\sin{x})'\\
\\
&= -\sin{x} \sin{x} + (1+\cos{x}) \times \cos{x}\\
\\
&= -\sin^2{x}+\cos{x}+\cos^2{x} \color{red}\scriptsize\ 　\sin,\ \cos どちらかに統一\\
& \color{red}\scriptsize　　　\Darr \sin^2{x}+\cos^2{x}=1\ より\ \cos^2{x} -1 = - \sin^2{x} \\
&= (\cos^2{x}-1)+\cos{x}+\cos^2{x}\\
&\\
&= 2\cos^2{x}+\cos{x}-1\\
\\
&= (2\cos{x}-1)(\cos{x}+1)
\end{align*}

\begin{align*}
f'(x) &= 0\\
(2\cos{x}-1)(\cos{x}+1) &= 0\\
\color{red}\scriptsize 2\cos{x}-1 = 0, & 　\color{red}\scriptsize\cos{x}+1=0\\
\cos{x} &= -1,\ \dfrac12\\
x &= \pi,\ \dfrac13\pi,\ \dfrac53\pi\\
\end{align*}

\def\arraystretch{1.8}
\begin{array}{c||c|c|c|c|c|c|c|c|c}\hline
x & 0 & \cdots & \frac{\pi}{3} & \cdots & \pi & \cdots & \frac53\pi & \cdots & 2\pi \\\hline
y' & \color{red}{\times} & + & 0 & - & 0 & - & 0 & + & \color{red}\times \\\hline
y & 0 & \nearrow & \colorbox{mistyrose}{$\substack{極大\\\frac{3\sqrt{3}}{4}}$} & \searrow & 0 & \searrow & \colorbox{lightcyan}{$\substack{極小\\-\frac{3\sqrt{3}}{4}}$} & \nearrow & 0 \\\hline
\end{array}

\begin{align*}
f(x) &= \dfrac{4-3x}{x^2+1}\color{red}\scriptsize\Leftarrow \frac{A}{B} \\
& \color{red}\scriptsize 　　　　A'　　　　　B　　-\ 　　A　　　　B'\\
f'(x) &= \dfrac{(4-3x)'(x^2+1)-(4-3x)(x^2+1)'}{(x^2+1)^2\color{red}\scriptsize\ \Leftarrow B^2}\\
\\
&= \dfrac{-3(x^2+1)-(4-3x) \cdot2x}{(x^2+1)^2}\\
\\
&= \dfrac{-3x^2-3-8x+6x^2}{(x^2+1)^2}\\
\\
&= \dfrac{3x^2-8x-3}{(x^2+1)^2}\\
\\
&= \dfrac{(3x+1)(x-3)}{(x^2+1)^2}\\
\end{align*}

\begin{align*}
f'(x) &= 0\\
\\
\dfrac{(3x+1)(x-3)}{(x^2+1)^2} &= 0\\
\color{red}\scriptsize 両辺\ (x^2+1)^2\ をかけて & \\
(3x+1)(x-3) &= 0\\
x &= -\dfrac13,\ 3\\
{\footnotesize 定義域}\ 1 \leqq x \leqq 4\ より &\\
x &= 3
\end{align*}

\def\arraystretch{1.8}
\begin{array}{c||c|c|c|c|c|}\hline
x & 1 & \cdots & 3 & \cdots & 4 \\\hline
y' & \color{red}\times & - & 0 & + & \color{red}\times\\\hline
y & \colorbox{mistyrose}{$\frac12$} & \searrow & \colorbox{lightcyan}{$\substack{極大\\-\frac12}$} & \nearrow & -\frac{8}{17}\\\hline
\end{array}