【数学I】例題3.2.8：おき換えを用いた最大・最小（One More）★★★

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% 例題I3.2.8：おき換えを用いた最大・最小（One More）★★★
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関数$y=(x^2-2x)^2+4(x^2-2x)$について，次の問いに答えよ．

(1) $t=x^2-2x$とおいて，$t$のとりうる値の範囲を求めよ．

(2) $y$を$t$の式で表し，$y$の最小値と，そのときの$x$の値を求めよ．

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(1)

\begin{align*}
t&=x^2-2x\\
&=(x-1)^2-1
\end{align*}

グラフは右の図のようになる．

よって，$t$のとりうる値の範囲は，$t\geqq-1$

(2) $t=x^2-2x$とおくと，

\[
y=t^2+4t=(t+2)^2-4\cdots(\mathrm{i})
\]

(1)より，$t\geqq-1$であるから，この範囲で，(i)のグラフをかくと，右の図のようになる．

したがって，$t=-1$のとき，$y$は最小値$-3$をとる．

また，$t=-1$のとき，$x^2-2x=-1$

ゆえに，$x=1$

よって，$x=1$のとき，最小値$-3$

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% 問題I3.2.8：おき換えを用いた最大・最小（One More）★★★
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関数$y=(x^2-4x)^2+6(x^2-4x)$について，次の問いに答えよ．

(1) $t=x^2-4x$とおいて，$t$のとりうる値の範囲を求めよ．

(2) $y$を$t$の式で表し，$y$の最小値と，そのときの$x$の値を求めよ．

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(1)

\begin{align*}
t&=x^2-4x\\
&=(x-2)^2-4
\end{align*}

グラフは右の図のようになる．

よって，$t$のとりうる値の範囲は，$t\geqq-4$

(2) $t=x^2-4x$とおくと，

\[
y=t^2+6t=(t+3)^2-9\cdots(\mathrm{i})
\]

(1)より，$t\geqq-4$であるから，この範囲で，(i)のグラフをかくと，右の図のようになる．

したがって，$t=-3$のとき，$y$は最小値$-9$をとる．

また，$t=-3$のとき，$x^2-4x=-3$

ゆえに，$x=1,3$

よって，$x=1,3$のとき，最小値$-9$

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% 側注1
$t$は$x$についての2次関数となるので，$x$軸，$y$軸ではなく，$x$軸，$t$軸であることに注意すること．

% 側注2
(1)で求めた$t$の値の範囲を忘れないようにする．また，$x$軸ではなく，$t$軸であることに注意すること．

% 側注3
$x^2-2x=-1$より，$(x-1)^2=0$となるので，これを解いて，$x=1$

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% 問題I3.2.8の側注：おき換えを用いた最大・最小（One More）★★★
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% 側注1
$t$は$x$についての2次関数となるので，$x$軸，$y$軸ではなく，$x$軸，{$t$軸}であることに注意すること．

% 側注2
(1)で求めた$t$の値の範囲を忘れないようにする．また，$x$軸ではなく，{$t$軸}であることに注意すること．

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% 例題I3.2.8の考え方・補足など：おき換えを用いた最大・最小（One More）★★★
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4次関数の問題であるが，共通する部分をおき換えることによって，2次関数の最大・最小の問題と見ることができる．$x^2-2x=t$とおくと，$y$は$t$の2次関数として考えることができる．なお，文字でおき換えた場合は，その文字の変域を考える必要があるので注意すること．

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共通する部分をおき換えて，2次関数と見て考える．このとき，おき換えた文字の変域に注意すること．