【数学I】例題3.2.5：軸が移動するときの最大・最小（One More）★★★

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% 例題I3.2.5：軸が移動するときの最大・最小（One More）★★★
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関数$f(x)=x^2-2ax+3(0\leqq x\leqq 3)$について，$f(x)$の最小値を求めよ．

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% 例題I3.2.5の解答（One More）★★★
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\[
y=x^2-2ax+3=(x-a)^2-a^2+3
\]

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=a$

(i) $a<0$のとき

グラフは右の図のようになり，軸は定義域より左側にある．

$x=0$のとき最小となり，最小値$f(0)=3$

(ii) $0\leqq a\leqq 3$のとき

グラフは右の図のようになり，軸は定義域内にある．

$x=a$のとき最小となり，最小値$f(a)=-a^2+3$

(iii) $3<a$のとき

グラフは右の図のようになり，軸は定義域より右側にある．

$x=3$のとき最小となり，最小値$f(3)=-6a+12$

よって，(i)〜(iii)より，

\[
\begin{cases}a<0\text{のとき，}&x=0\text{で最小値}3\\0\leqq a\leqq 3\text{のとき，}&x=a\text{で最小値}-a^2+3\\3<a\text{のとき，}&x=3\text{で最小値}-6a+12\end{cases}
\]

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% 問題I3.2.5：軸が移動するときの最大・最小（One More）★★★
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(1) 関数$f(x)=x^2-2ax+3(0\leqq x\leqq 3)$について，$f(x)$の最大値を求めよ．

(2) 関数$f(x)=x^2-2ax+5(1\leqq x\leqq 4)$について，$f(x)$の最小値を求めよ．

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% 問題I3.2.5の解答（One More）★★★
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(1)

$y=x^2-2ax+3=(x-a)^2-a^2+3$

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=a$

(i) $a<\frac{3}{2}$のとき

グラフは右の図のようになる．

$x=3$のとき最大となり，最大値$f(3)=-6a+12$

(ii) $a=\frac{3}{2}$のとき

グラフは右の図のようになる．

$x=0,3$のとき最大となり，最大値$f(0)=f(3)=3$

(iii) $\frac{3}{2}<a$のとき

グラフは右の図のようになる．

$x=0$のとき最大となり，最大値$f(0)=3$

よって，(i)〜(iii)より，

\[
\begin{cases}a<\frac{3}{2}\text{のとき，}&x=3\text{で最大値}-6a+12\\a=\frac{3}{2}\text{のとき，}&x=0,3\text{で最大値}3\\\frac{3}{2}<a\text{のとき，}&x=0\text{で最大値}3\end{cases}
\]

(2)

$y=x^2-2ax+5=(x-a)^2-a^2+5$

$y=f(x)$のグラフは下に凸の放物線で，軸は直線$x=a$

(i) $a<1$のとき

グラフは右の図のようになり，軸は定義域より左側にある．

$x=1$のとき最小となり，最小値$f(1)=-2a+6$

(ii) $1\leqq a\leqq 4$のとき

グラフは右の図のようになり，軸は定義域内にある．

$x=a$のとき最小となり，最小値$f(a)=-a^2+5$

(iii) $4<a$のとき

グラフは右の図のようになり，軸は定義域より右側にある．

$x=4$のとき最小となり，最小値$f(4)=-8a+21$

よって，(i)〜(iii)より，

\[
\begin{cases}a<1\text{のとき，}&x=1\text{で最小値}-2a+6\\1\leqq a\leqq 4\text{のとき，}&x=a\text{で最小値}-a^2+5\\4<a\text{のとき，}&x=4\text{で最小値}-8a+21\end{cases}
\]

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% 例題I3.2.5の側注：軸が移動するときの最大・最小（One More）★★★
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% 側注1
軸が定義域より左側か右側にあるとき，定義域の左端か右端で最小値をとる．これと軸が定義域内にある場合，頂点で最小値をとることから，全部で3通りの場合分けとなる．

% 側注2
頂点で最小値をとる．

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% 問題I3.2.5の側注：軸が移動するときの最大・最小（One More）★★★
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% 側注1
軸が定義域の中央である$x=\frac{3}{2}$より左側にあるとき，右側にあるときで場合分けをする．

% 側注2
$x=0$の方が軸から遠いので，$x=0$で最大となる．

% 側注3
頂点で最小値をとる．

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% 例題I3.2.5の考え方・補足など：軸が移動するときの最大・最小（One More）★★★
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% 考え方
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関数$f(x)$を標準形にすると，$f(x)=(x-a)^2-a^2+3$となり，軸は直線$x=a$である．$a$の値が変わると，軸の位置も変わり，定義域における最小値をとる点（端点または頂点）も変わる．よって，定義域に対する軸の位置を考えて$a$について場合分けをする．

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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) $a<0$

(ii) $0\leqq a<\frac{3}{2}$

(iii) $a=\frac{3}{2}$

(iv) $\frac{3}{2}<a\leqq 3$

(v) $3<a$