【数学I】例題3.2.3：最大・最小による係数の決定（One More）★★★

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% 例題I3.2.3：最大・最小による係数の決定（One More）★★★
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関数$f(x)=ax^2-4ax+2b(-1\leqq x\leqq 3)$の最大値が$7$，最小値が$-2$のとき，定数$a,b$の値を求めよ．

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% 例題I3.2.3の解答（One More）★★★
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\[
f(x)=ax^2-4ax+2b=a(x-2)^2-4a+2b
\]

(i) $a>0$のとき

$y=f(x)$のグラフは下に凸の放物線で，$-1\leqq x\leqq 3$の範囲で$f(x)$は，$x=-1$のとき，最大値$5a+2b$，$x=2$のとき，最小値$-4a+2b$をとる．

したがって，$\{\begin{array}{l}5a+2b=7\\-4a+2b=-2\end{array}$

ゆえに，$a=1,b=1$

これは$a>0$を満たす．

(ii) $a=0$のとき

$f(x)=2b$で一定の値をとり，最大値7，最小値$-2$をとることはないから，不適である．

(iii) $a<0$のとき

$y=f(x)$のグラフは上に凸の放物線で，$-1\leqq x\leqq 3$の範囲で$f(x)$は，$x=2$のとき，最大値$-4a+2b$，$x=-1$のとき，最小値$5a+2b$をとる．

したがって，$\{\begin{array}{l}-4a+2b=7\\5a+2b=-2\end{array}$

ゆえに，$a=-1,b=\frac{3}{2}$

これは$a<0$を満たす．

よって，(i)〜(iii)より，求める$a,b$の値は，$(a,b)=(1,1),(-1,\frac{3}{2})$

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% 問題I3.2.3：最大・最小による係数の決定（One More）★★★
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関数$f(x)=ax^2-4ax+2b(-1\leqq x\leqq 2)$の最大値が$5$，最小値が$-1$のとき，定数$a,b$の値を求めよ．

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% 問題I3.2.3の解答（One More）★★★
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\[
f(x)=ax^2-4ax+2b=a(x-2)^2-4a+2b
\]

(i) $a>0$のとき

$y=f(x)$のグラフは下に凸の放物線で，$-1\leqq x\leqq 2$の範囲で$f(x)$は，$x=-1$のとき，最大値$5a+2b$，$x=2$のとき，最小値$-4a+2b$をとる．

したがって，$\{\begin{array}{l}5a+2b=5\\-4a+2b=-1\end{array}$

ゆえに，$a=\frac{2}{3},b=\frac{5}{6}$

これは$a>0$を満たす．

(ii) $a=0$のとき

$f(x)=2b$で一定の値をとり，最大値$5$，最小値$-1$をとることはないから，不適である．

(iii) $a<0$のとき

$y=f(x)$のグラフは上に凸の放物線で，$-1\leqq x\leqq 2$の範囲で$f(x)$は，$x=2$のとき，最大値$-4a+2b$，$x=-1$のとき，最小値$5a+2b$をとる．

したがって，$\{\begin{array}{l}-4a+2b=5\\5a+2b=-1\end{array}$

ゆえに，$a=-\frac{2}{3},b=\frac{7}{6}$

これは$a<0$を満たす．

よって，(i)〜(iii)より，求める$a,b$の値は，$(a,b)=(\frac{2}{3},\frac{5}{6}),(-\frac{2}{3},\frac{7}{6})$

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% 例題I3.2.3の側注：最大・最小による係数の決定（One More）★★★
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% 側注1
軸は直線$x=2$，頂点は点$(2,-4a+2b)$の放物線である．

% 側注2
軸から遠い方の端点である$x=-1$のとき，最大となる．

% 側注3
場合分けの条件を満たすことを確認する．

% 側注4
軸から遠い方の端点である$x=-1$のとき，最小となる．

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% 問題I3.2.3の側注：最大・最小による係数の決定（One More）★★★
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% 側注1
軸から遠い方の端点である$x=-1$のとき，最大となる．

% 側注2
場合分けの条件を満たすことを確認する．

% 側注3
軸から遠い方の端点である$x=-1$のとき，最小となる．

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% 例題I3.2.3の考え方・補足など：最大・最小による係数の決定（One More）★★★
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% 考え方
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$a>0\text{（下に凸の放物線）},a=0\text{（直線）},a<0\text{（上に凸の放物線）}$の3つの場合に分けて考える．なお，場合分けで得られた値が，場合分けの条件を満たすか否かの確認を忘れないように注意すること．

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% 注意
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問題文では単に「関数$f(x)$」となっていることから，$f(x)$は2次関数であるとは限らない．「2次関数$f(x)$」であれば$a\neq 0$は仮定されていると考えるが，「関数$f(x)$」となっていることから，$a=0$のときを考えなければならないので注意すること．