【数学I】例題3.2.2：定義域が定められたときの2次関数の最大・最小（One More）★★

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% 例題I3.2.2：定義域が定められたときの2次関数の最大・最小（One More）★★
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次の定義域における2次関数$y=x^2-6x+5$の最大値，最小値を求めよ．

(1) $1\leqq x\leqq 4$

(2) $0<x\leqq 2$

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% 例題I3.2.2の解答（One More）★★
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\[
y=x^2-6x+5=\{(x-3)^2-3^2\}+5=(x-3)^2-4
\]

(1) $x=1$のとき，$y=0$，$x=4$のとき，$y=-3$

したがって，$1\leqq x\leqq 4$のとき，グラフは右の図のようになる．

よって，$x=1$のとき，最大値$0$，$x=3$のとき，最小値$-4$

(2) $x=0$のとき，$y=5$，$x=2$のとき，$y=-3$

したがって，$0<x\leqq 2$のとき，グラフは右の図のようになる．

よって，$x=2$のとき，最小値$-3$，最大値はない．

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% 問題I3.2.2：定義域が定められたときの2次関数の最大・最小（One More）★★
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次の定義域における2次関数$y=-x^2+4x+1$の最大値，最小値を求めよ．

(1) $0\leqq x<2$

(2) $1<x\leqq 4$

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% 問題I3.2.2の解答（One More）★★
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\[
y=-x^2+4x+1=-\{(x-2)^2-4\}+1=-(x-2)^2+5
\]

(1) $x=0$のとき，$y=1$，$x=2$のとき，$y=5$

したがって，$0\leqq x<2$のとき，グラフは右の図のようになる．

よって，$x=0$のとき，最小値$1$，最大値はない．

(2) $x=1$のとき，$y=4$，$x=4$のとき，$y=1$

したがって，$1<x\leqq 4$のとき，グラフは右の図のようになる．

よって，$x=2$のとき，最大値$5$，$x=4$のとき，最小値$1$

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% 例題I3.2.2の側注：定義域が定められたときの2次関数の最大・最小（One More）★★
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% 側注1
$x=0$を代入すると，$y=0^2-6\cdot 0+5=5$，$x=2$を代入すると，$y=2^2-6\cdot 2+5=-3$

% 側注2
$0<x\leqq 2$より，$x=0$は定義域に含まれないので，最大値はない．

% 側注3
グラフは下に凸の放物線で，軸は直線$x=3$，頂点は点$(3,-4)$である．

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% 問題I3.2.2の側注：定義域が定められたときの2次関数の最大・最小（One More）★★
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% 側注1
$0\leqq x<2$より，$x=2$は定義域に含まれないので，最大値はない．

% 側注2
平方完成する．

% 側注3
グラフは上に凸の放物線で，軸は直線$x=2$，頂点は点$(2,5)$である．

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% 例題I3.2.2の考え方・補足など：定義域が定められたときの2次関数の最大・最小（One More）★★
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% 考え方
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2次関数の最大・最小には，グラフの利用が有効である．このとき，定義域が定められている場合は，軸が定義域に含まれるか否かに注意すること．放物線は軸で対称であることを利用して，軸と定義域の端の値（端点）の位置関係から，最大値・最小値を求めることができる．

軸が定義域に含まれるとき

軸が定義域に含まれないとき

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定義域が定められた2次関数の最大・最小は，軸（頂点）の位置と定義域の端の値に注目する．