【数学I】例題3.1.11：絶対値記号を含む関数のグラフ2（One More）★★★

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% 例題I3.1.11：絶対値記号を含む関数のグラフ2（One More）★★★
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関数$y=|x+3|+|x-1|$のグラフをかけ．

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% 例題I3.1.11の解答（One More）★★★
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(i) $x<-3$のとき

\begin{align*}
|x+3|+|x-1|&=-(x+3)-(x-1)\\
&=-2x-2
\end{align*}

(ii) $-3\leqq x<1$のとき

\begin{align*}
|x+3|+|x-1|&=(x+3)-(x-1)\\
&=4
\end{align*}

(iii) $1\leqq x$のとき

\begin{align*}
|x+3|+|x-1|&=(x+3)+(x-1)\\
&=2x+2
\end{align*}

よって，(i)〜(iii)より，グラフは右上の図のようになる．

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% 問題I3.1.11：絶対値記号を含む関数のグラフ2（One More）★★★
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関数$y=|x+2|+|x-4|$のグラフをかけ．

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% 問題I3.1.11の解答（One More）★★★
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(i) $x<-2$のとき

\begin{align*}
|x+2|+|x-4|&=-(x+2)-(x-4)\\
&=-2x+2
\end{align*}

(ii) $-2\leqq x<4$のとき

\begin{align*}
|x+2|+|x-4|&=(x+2)-(x-4)\\
&=6
\end{align*}

(iii) $x\geqq 4$のとき

\begin{align*}
|x+2|+|x-4|&=(x+2)+(x-4)\\
&=2x-2
\end{align*}

よって，(i)〜(iii)より，グラフは右上の図のようになる．

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% 例題I3.1.11の側注：絶対値記号を含む関数のグラフ2（One More）★★★
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% 側注1
定数関数$y=4$となる．

% 側注2
グラフは繋がっている．

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% 問題I3.1.11の側注：絶対値記号を含む関数のグラフ2（One More）★★★
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% 側注1
定数関数$y=6$となる．

% 側注2
グラフは繋がっている．

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% 例題I3.1.11の考え方・補足など：絶対値記号を含む関数のグラフ2（One More）★★★
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% 考え方
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絶対値記号内が$0$になる値を境に場合分けをする．$x+3=0$より，$x=-3$，$x-1=0$より，$x=1$であるから，$x=-3,1$を境に，$x<-3,-3\leqq x<1,1\leqq x$の3つの部分に分けて場合分けをする．

\[
\begin{aligned}
&|x+3|=\begin{cases}x+3&(x\geqq-3)\\
-(x+3)&(x<-3)\end{cases}
\end{aligned}
\begin{aligned}
,|x-1|=\{\begin{array}{ll}
x-1&(x\geqq 1)\\
-(x-1)&(x<1)
\end{array}
\end{aligned}
\]

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% 余談
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一般に，$y=|f(x)|,y=|f(x)|\pm|g(x)|$（$f(x),g(x)$は1次式や2次式などの多項式とする）などの形の関数は，場合分けの境となる$x$でグラフが繋がっていることが知られている．

グラフが繋がっている

グラフが繋がっている