【数学I】例題3.2.6：定義域が変化するときの最大・最小（One More）★★★

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% 例題I3.2.6：定義域が変化するときの最大・最小（One More）★★★
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関数$f(x)=x^2-2x+3(a\leqq x\leqq a+2)$について，$f(x)$の最大値を求めよ．

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% 例題I3.2.6の解答（One More）★★★
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\[
f(x)=x^2-2x+3=(x-1)^2+2
\]

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=1$

(i) $a+1<1$，すなわち，$a<0$のとき

グラフは右の図のようになる．

$x=a$で最大となり，最大値$f(a)=a^2-2a+3$

(ii) $a+1=1$，すなわち，$a=0$のとき

グラフは右の図のようになる．

$x=0,2$で最大となり，最大値$f(0)=f(2)=3$

(iii) $1<a+1$，すなわち，$a>0$のとき

グラフは右の図のようになる．

$x=a+2$で最大となり，最大値$f(a+2)=a^2+2a+3$

よって，(i)〜(iii)より，

\[
\begin{cases}a<0\text{のとき，}&x=a\text{で最大値}a^2-2a+3\\a=0\text{のとき，}&x=0,2\text{で最大値}3\\a>0\text{のとき，}&x=a+2\text{で最大値}a^2+2a+3\end{cases}
\]

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% 問題I3.2.6：定義域が変化するときの最大・最小（One More）★★★
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(1) 関数$f(x)=x^2-2x+3(a\leqq x\leqq a+2)$について，$f(x)$の最小値を求めよ．

(2) 関数$f(x)=x^2-4x+5(a\leqq x\leqq a+2)$について，$f(x)$の最大値を求めよ．

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% 問題I3.2.6の解答（One More）★★★
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(1)

$f(x)=x^2-2x+3=(x-1)^2+2$

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=1$

(i) $a+2<1$，すなわち，$a<-1$のとき

グラフは右の図のようになる．

$x=a+2$で最小となり，最小値$f(a+2)=a^2+2a+3$

(ii) $a\leqq 1\leqq a+2$，すなわち，$-1\leqq a\leqq 1$のとき

グラフは右の図のようになる．

$x=1$で最小となり，最小値$f(1)=2$

(iii) $1<a$のとき

グラフは右の図のようになる．

$x=a$で最小となり，最小値$f(a)=a^2-2a+3$

よって，(i)〜(iii)より，

\[
\begin{cases}a<-1\text{のとき，}&x=a+2\text{で最小値}a^2+2a+3\\
-1\leqq a\leqq 1\text{のとき，}&x=1\text{で最小値}2\\
1<a\text{のとき，}&x=a\text{で最小値}a^2-2a+3\end{cases}
\]

(2)

$f(x)=x^2-4x+5=(x-2)^2+1$

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=2$

(i) $a+1<2$，すなわち，$a<1$のとき

グラフは右の図のようになる．

$x=a$で最大となり，最大値$f(a)=a^2-4a+5$

(ii) $a+1=2$，すなわち，$a=1$のとき

グラフは右の図のようになる．

$x=1,3$で最大となり，最大値$f(1)=f(3)=2$

(iii) $2<a+1$，すなわち，$a>1$のとき

グラフは右の図のようになる．

$x=a+2$で最大となり，最大値$f(a+2)=a^2+1$

よって，(i)〜(iii)より，

\[
\begin{cases}a<1\text{のとき，}&x=a\text{で最大値}a^2-4a+5\\a=1\text{のとき，}&x=1,3\text{で最大値}2\\a>1\text{のとき，}&x=a+2\text{で最大値}a^2+1\end{cases}
\]

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% 例題I3.2.6の側注：定義域が変化するときの最大・最小（One More）★★★
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% 側注1
定義域$a\leqq x\leqq a+2$の中央である$x=a+1$と軸の直線$x=1$が一致するときを境にして，場合分けをする．ここでは，$a+1=1$，すなわち，$a=0$を境に場合分けをする．また，(i)は$x=a$の方が軸から遠いので，$x=a$で最大となる．

% 側注2
$x=a+2$の方が軸から遠いので，$x=a+2$で最大となる．

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% 問題I3.2.6の側注：定義域が変化するときの最大・最小（One More）★★★
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% 側注1
軸が定義域の左側にあるとき，区間内にあるとき，右側にあるときで場合分けをする．

% 側注2
$x=a$の方が軸から遠いので，$x=a$で最大となる．

% 側注3
$x=a+2$の方が軸から遠いので，$x=a+2$で最大となる．

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% 例題I3.2.6の考え方・補足など：定義域が変化するときの最大・最小（One More）★★★
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% 考え方
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定義域が変化するが，定義域の幅は常に$2$で一定である．定義域の中央と軸に注目して，場合分けをする．定義域の中央は，$\frac{a+(a+2)}{2}=a+1$であり，これと軸$x=1$が一致するとき，すなわち，$a+1=1$より$a=0$で定義域の両端と軸までの距離が等しくなる．よって，$a=0$を境に場合分けをすればよい．

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% 注意
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下の問題の(1)の結果と合わせると，最大値と最小値は次のようにまとめることができる．

(i) $a<-1$

(ii) $-1\leqq a<0$

(iii) $a=0$

(iv) $0<a\leqq 1$

(v) $1<a$