微分して，関数の増減を調べよう

\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\begin{align*}
f'(x) = 3x^2-3 　\ \ \  &\\
\\
\colBX{bisque}{$f'(x)=0$} とすると\\
3x^2-3 &= 0\\
3x^2 &= 3\\
x^2 &= 1\\
x &= \colBX{palegreen}{$-1$},\ \colBX{palegreen}{$1$}\\
\\
\colBX{violet}{$f'(x)>0$} とすると\\
3x^2-3 &> 0\\
x^2-1 &> 0\\
(x+1)(x-1) &> 0\\
x<\colBX{palegreen}{$-1$},\ \colBX{palegreen}{$1$} &< x\\
\\
\colBX{lightcyan}{$f'(x)<0$} とすると\\
3x^2-3 &< 0\\
x^2-1 &< 0\\
(x+1)(x-1) &< 0\\
\colBX{palegreen}{$-1$} < x &< \colBX{palegreen}{$1$}
\end{align*}

\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\def\arraystretch{1.2}
\begin{array}{c||c|c|c|c|c}\hline
x & \cdots & \colBX{palegreen}{$-1$} & \cdots & \colBX{palegreen}{$1$} & \cdots \\\hline
f'(x) & \colBX{violet}{$+$} & \colBX{bisque}{$0$} & \colBX{lightcyan}{$-$} & \colBX{bisque}{$0$} & \colBX{violet}{$+$}\\\hline
f(x) & \colBX{violet}{$\nearrow$} & 2 & \colBX{lightcyan}{$\searrow$} & -2 & \colBX{violet}{$\nearrow$}\\\hline
\end{array}

\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\begin{align*}
したがって,\ f(x)\ は\\
\\
\colBX{violet}{$x \leqq -1$},\ \colBX{violet}{$1 \leqq x$} &\ で増加し\\
\\
\colBX{lightcyan}{$-1 \leqq x \leqq 1$} &\ で減少する
\end{align*}